To find the length of border needed to put around a rectangular room, you can follow these steps:
Measure the length and width of the room using a tape measure or ruler. Let's say the length is L and the width is W.
Determine the type and width of the border you want to put around the room. Let's say you want to use a wallpaper border that is 6 inches wide.
Calculate the length of the border needed for the top and bottom of the room.
This can be done by adding the width of the room to twice the width of the border. So the length of the top and bottom borders would be:
Length of top/bottom borders = 2(W + 6 inches)
Calculate the length of the border needed for the left and right sides of the room. This can be done by adding the length of the room to twice the width of the border. So the length of the left and right borders would be:
Length of left/right borders = 2(L + 6 inches)
Add up the length of all four borders to get the total length of border needed to put around the room:
Total length of border = 2(W + 6 inches) + 2(L + 6 inches)
Simplifying the above equation, we get:
Total length of border = 2L + 2W + 24 inches
So the total length of border needed to put around the room is 2L + 2W + 24 inches.
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Find the line of tangency to the circle defined by (x-3)^2 + (y-7)^2 = 169 at the point (15,2).
first off, let's look at the equation of the circle
[tex]\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{}{h}~~,~~\underset{}{k})}\qquad \stackrel{radius}{\underset{}{r}} \\\\[-0.35em] ~\dotfill\\\\ (x-\stackrel{h}{3})^2+(y-\stackrel{k}{7})=169\implies (x-\stackrel{h}{3})^2+(y-\stackrel{k}{7})=\stackrel{ r }{13^2}[/tex]
so we have a circle centered at (3 , 7) with a radius of 13, Check the picture below.
so the line we want is the line in purple, which is tangential to the circle and therefore perpendicular to the blue line.
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the blue line
[tex](\stackrel{x_1}{3}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{7}}}{\underset{\textit{\large run}} {\underset{x_2}{15}-\underset{x_1}{3}}} \implies \cfrac{ -5 }{ 12 } \implies - \cfrac{5 }{ 12 } \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{-5}{12}} ~\hfill \stackrel{reciprocal}{\cfrac{12}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{12}{-5} \implies \cfrac{12}{ 5 }}}[/tex]
so we're really looking for the equation of a line whose slope is 12/5 and it passes through (15 , 2)
[tex](\stackrel{x_1}{15}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{12}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{ \cfrac{12}{5}}(x-\stackrel{x_1}{15}) \\\\\\ y-2=\cfrac{12}{5}x-36\implies {\Large \begin{array}{llll} y=\cfrac{12}{5}x-34 \end{array}}[/tex]
If $5,500 is invested at 1.55% interest, find the value (in dollars) of the investment at the end of 6 years if the interest is
compounded as follows. Roung vour answers to the nearest cent.
A. Annualy
B. quarterly
C. Monthly
The value of the investment at the end of 6 years is $6,359.77 if the interest is compounded annually, $6,416.52 if it is compounded quarterly, and $6,437.70 if it is compounded monthly.
What formula is used to for compound interest?
[tex]A = P(1 + r/n)^{nt}[/tex]
where A is the final amount, P is the initial investment, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.
A. If the interest is compounded annually, we have:
A = [tex]5,500(1 + 0.0155/1)^{1*6}[/tex] = $6,359.77
B. If the interest is compounded quarterly, we have:
A = [tex]5,500(1 + 0.0155/4)^{4*6[/tex] = $6,416.52
C. If the interest is compounded monthly, we have:
A = [tex]5,500(1 + 0.0155/12)^{12*6[/tex] = $6,437.70
Therefore, the value of the investment at the end of 6 years is $6,359.77 if the interest is compounded annually, $6,416.52 if it is compounded quarterly, and $6,437.70 if it is compounded monthly.
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if x varies directly as T and x=105 when T=400, find x when T=500
Answerx = 131.25
Step-by-step explanation:f x varies directly as T, then we can use the formula for direct variation:
x = kT
where k is the constant of proportionality.
To find k, we can use the given values:
x = 105 when T = 400
105 = k(400)
k = 105/400
k = 0.2625
Now that we have the value of k, we can use the formula to find x when T = 500:
x = kT
x = 0.2625(500)
x = 131.25
Therefore, when T = 500, x is equal to 131.25.
A rectangular brick wall is 6 wide and 1 m tall. Use Pythagoras' theorem to work out the distance between diagonally opposite corners. Give your answer in metres (m) to 1 d.p.
Pythagorean Theorem: a^2 + b^2 = c^2
---a and b are the legs of the triangle
---c is the hypotenuse/diagonal
a = 6
b = 1
c = ?
(6)^2 + (1)^2 = c^2
36 + 1 = c^2
37 = c^2
c = 6.0827
c (rounded) = 6.1
Answer = 6.1 meters
Internet Use A survey of U. S. Adults ages 18–29 found that 93% use the
Internet. You randomly select 100 adults ages 18–29 and ask them if they use
the Internet.
(a) Find the probability that exactly 90 people say they use the Internet.
(b) Find the probability that at least 90 people say they use the Internet.
(c) Find the probability that fewer than 90 people say they use the Internet.
(d) Are any of the probabilities in parts (a)-(c) unusual? Explain.
a. The probability that exactly 90 people say they use the Internet is 0.0391
b. The probability that at least 90 people say they use the Internet is 0.1933
c. The probability that fewer than 90 people say they use the Internet is 0.8067
d. The first probability is unusual
How to solve the problems(a) Find the probability that exactly 90 people say they use the Internet.
P(X = 90) = C(100, 90) * (0.93)^90 * (0.07)^10
P(X = 90) ≈ 0.0391
(b) Find the probability that at least 90 people say they use the Internet.
P(X ≥ 90) = P(X = 90) + P(X = 91) + ... + P(X = 100)
To calculate this, we can use cumulative binomial probability:
P(X ≥ 90) ≈ 1 - P(X ≤ 89) ≈ 1 - 0.8067 = 0.1933
(c) Find the probability that fewer than 90 people say they use the Internet.
P(X < 90) = P(X ≤ 89)
P(X < 90) ≈ 0.8067
(d) Are any of the probabilities in parts (a)-(c) unusual? Explain.
A probability is generally considered unusual if it is less than 0.05 or greater than 0.95. Based on the calculated probabilities:
P(X = 90) ≈ 0.0391: This probability is unusual since it is less than 0.05.
P(X ≥ 90) ≈ 0.1933: This probability is not unusual.
P(X < 90) ≈ 0.8067: This probability is not unusual.
So, only the probability of exactly 90 people saying they use the Internet is considered unusual.
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Show that the two straight lines through the origin which make an angle 45° with the line px + qy + r = 0 are given by the equation (p²-q)(x² - y²) + 4pqxy = 0.
The equation of the two straight lines through the origin making an angle of 45° with the line px + qy + r = 0 is (p²-q)(x² - y²) + 4pqxy = 0.
How to show the equation for the two straight lines passing through the origin and making a 45° angle with the line px + qy + r = 0?To prove that the equation of the two straight lines through the origin, making an angle of 45° with the line px + qy + r = 0, is given by (p²-q)(x² - y²) + 4pqxy = 0, we can use the concept of slopes and trigonometric identities.
Let's consider the line px + qy + r = 0. The slope of this line is given by -p/q.
Now, the lines making an angle of 45° with this line will have slopes equal to tan(45°), which is 1.
Using the formula for the tangent of the sum of angles, we have:
tan(45°) = (m - (-p/q))/(1 + m(-p/q)), where m represents the slope of one of the lines.
Simplifying the equation, we get:
1 = (mq + p)/(q - mp)
Cross-multiplying and rearranging the terms, we obtain:
(p² - q)(m² - 1) + 2pqm = 0
Since these lines pass through the origin (0,0), we can replace m with y/x. Substituting y/x for m in the equation above, we get:
(p² - q)(x² - y²) + 2pqxy = 0
Further simplifying the equation, we arrive at:
(p² - q)(x² - y²) + 4pqxy = 0
Hence, we have proven that the equation of the two straight lines through the origin, making an angle of 45° with the line px + qy + r = 0, is given by (p²-q)(x² - y²) + 4pqxy = 0.
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Find the values of a and b, if the function defined by f(x) = x^2 + 3x + a , x <= 1
bx + 2, x >= 1 is differentiable at x = 1
To find the values of a and b, we need to ensure that the function is differentiable at x = 1. Thus, the function defined by f(x) = x^2 + 3x + a, x <= 1 and bx + 2, x >= 1 differentiable at x = 1 are a = 3 and b = 5.
First, we need to check that the function is continuous at x = 1. Since the function has different definitions for x <= 1 and x >= 1, we need to check that the limit of the function as x approaches 1 from both sides is the same.
Limit as x approaches 1 from the left (x <= 1):
f(x) = x^2 + 3x + a
lim x->1- f(x) = lim x->1- (x^2 + 3x + a) = 1^2 + 3(1) + a = 4 + a
Limit as x approaches 1 from the right (x >= 1):
f(x) = bx + 2
lim x->1+ f(x) = lim x->1+ (bx + 2) = b + 2
For the function to be continuous at x = 1, these two limits must be equal.
4 + a = b + 2
a = b - 2
Now we need to check that the derivative of the function at x = 1 exists and is equal from both sides.
Derivative of the function for x <= 1:
f(x) = x^2 + 3x + a
f'(x) = 2x + 3
f'(1) = 2(1) + 3 = 5
Derivative of the function for x >= 1:
f(x) = bx + 2
f'(x) = b
f'(1) = b
For the function to be differentiable at x = 1, these two derivatives must be equal.
5 = b
Substituting b = 5 into the equation we found earlier for a, we get:
a = 5 - 2 = 3
Therefore, the values of a and b that make the function defined by f(x) = x^2 + 3x + a, x <= 1 and bx + 2, x >= 1 differentiable at x = 1 are a = 3 and b = 5.
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bruce's weigh 11 each if bruce has 14 how many pounds do they weigh all together
All the Bruces together weigh a total of 154 pounds.
If each Bruce weighs 11 pounds and there are 14 Bruces, we can calculate the total weight by multiplying the weight of each Bruce by the number of Bruces.
Weight of each Bruce: 11 pounds
Number of Bruces: 14
To find the total weight, we multiply 11 pounds by 14 Bruces:
Total weight = 11 pounds/Bruce * 14 Bruces
= 154 pounds
Therefore, all the Bruces together weigh a total of 154 pounds.
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What is the ratio of hours spent at soccer practice to hours spent at a birthday party? choose 1 answer:
The ratio of hours spent at soccer practice to hours spent at a birthday party can be represented as 2 for every 3
To provide an accurate ratio, I would need the specific number of hours spent at both soccer practice and the birthday party.
Once you provide that information, you can create the ratio by putting the two numbers in the form 2 for every 3.
For example, if you spent 3 hours at soccer practice and 2 hours at a birthday party, the ratio would be 3:2. This means that for every 3 hours spent at soccer practice, you spent 2 hours at the birthday party.
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The table shows how many hours Sara spent at several activities one Saturday.
Activity "Hours
Soccer practice 2
Birthday party 3
Science project 1
What is the ratio of hours spent at soccer practice to hours spent at a birthday party?
Choose 1 answer:
1 for every 2
B
2 for every 1
2 for every 3
3 for every 2
Determine whether the given points represent the vertices of a trapezoid If so, determine whether it is isoscoles or not
A(-4,-1),B((-4,6),C(2,6),D(2,-4)
Answer:
It is a trapezoid
Step-by-step explanation:
Yes, the given points represent the vertices of a trapezoid.
A trapezoid is a quadrilateral with one set of parallel sides. In this case, the parallel sides are AB and CD. The other two sides, AD and BC, are not parallel.
The trapezoid is not isosceles because the two non-parallel sides are not congruent. AD has a length of 6 units, while BC has a length of 4 units.
Here is a diagram of the trapezoid:
A(-4,-1)
B((-4,6)
C(2,6)
D(2,-4)
A trapezoid is a quadrilateral with at least one pair of parallel sides. In this case, sides AB and CD are parallel because they have the same slope. So, the given points do represent the vertices of a trapezoid.
An isosceles trapezoid has two congruent legs (non-parallel sides). In this case, the length of side AD is `sqrt((-4-2)^2+(-1+4)^2)=sqrt(36+9)=sqrt(45)` and the length of side BC is `sqrt((-4-2)^2+(6-6)^2)=sqrt(36+0)=sqrt(36)`. Since `sqrt(45)` is not equal to `sqrt(36)`, the trapezoid is not isosceles.
Solve x∕3 < 5 Question 12 options: A) x < 15 B) x ≥ 15 C) x ≤ 15 D) x > 15
Answer:
A) x < 15
Step-by-step explanation:
You want the solution to x/3 < 5.
InequalityThe steps to solving an inequality are basically identical to the steps for solving an equation. There are a couple of differences:
the direction of the inequality symbol must be respectedmultiplication/division by negative numbers reverses the inequality symbol1-stepIf this were and equation, it would be a "one-step" equation. That step is to multiply both sides by the inverse of the coefficient of x.
The coefficient of x is 1/3. Its inverse is 3. Multiplying both sides by 3, we have ...
3(x/3) < 3(5)
x < 15 . . . . . . . . . simplify
Note that 3 is a positive number, so we leave the inequality symbol pointing the same direction.
__
Additional comment
We can swap the sides of an equation based on the symmetric property of equality:
a = b ⇔ b = a
When we swap the sides of an inequality, we need to preserve the relationship between them. (This is the meaning of "respect the direction of the inequality symbol".)
a < b ⇔ b > a
Besides multiplying and dividing by a negative number, there are other operations that affect the order of values.
-2 < 1 ⇔ 2 > -1 . . . . . multiply by -12 < 3 ⇔ 1/2 > 1/3 . . . . . take the reciprocal (same signs)a < b ⇔ cot⁻¹(a) > cot⁻¹(b) . . . . use function having negative slopeNote that the 1/x function is another one that has negative slope, which is why it reverses the ordering for values with the same sign. (It has no effect on ordering of values with opposite signs.)
I Need help with this math problem
The value of angle x = 114°.
How to find angle x?From the figure, it is clear that The interior angle of a triangle is 39°, by the law of opposite angle.
The sum of the interior angle of a triangle is 180°
37° + 39° + ∠unknown1 = 180°
∠unkonown1 = 180° - 37° - 39°
∠unknown1 = 104°
The sum of the exterior angle and the interior angle is 180°.
∠unknown2+ ∠unknown 1= 180°
∠unknown2 = 180° - 104°
∠unknown2 = 76°
The sum of the interior angle of a triangle is 180°
∠unknown3 + ∠unknown2 + 38 = 180
∠unknown3 + 76° + 38 = 180
∠unknown3= 66°
The sum of the exterior angle and the interior angle is 180°.
∠X + <unknown3 = 180°
∠X = 180° - 66°
∠X = 114°
The value of the angle x is 114°.
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Triangle GHC is similar to triangle JKC what is the length of GC?
The length of GC is 22.5
Here we know that the triangles GHC and JKC are similar, we know that the corresponding angles are congruent.
That is, angle GHC is equal to angle JKC, angle HGC is equal to angle KJC, and angle CHG is equal to angle CKJ.
∠GHC = ∠JKC
∠HGC = ∠KJC
∠CHG = ∠CKJ
We can denote this as:
ΔGHC ~ Δ JKC
Using the concept of similarity, we can write the following proportion:
GC/ JK = HC/ KC
Here, GC represents the length of the corresponding side of triangle GHC, and JK represents the length of the corresponding side of triangle JKC. HC and KC represent the lengths of the other sides that are also proportional.
We can cross-multiply to get:
GC × KC = HC × JK
Now, we can substitute the values we know.
=> x * 20 = 18 x 25
=> x = 22.5
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Using Newton's Method, estimate the positive solution to the following equation by calculating x2 and using X0 = 1. x⁴ – x = 3 Round to four decimal places.
Answer:
To estimate the positive solution to the equation x⁴ – x = 3 using Newton's Method, we can start by taking the derivative of the equation, which is 4x³ - 1. Then we can use the formula X1 = X0 - f(X0) / f'(X0), where X0 = 1, f(X0) = 1⁴ - 1 - 3 = -3, and f'(X0) = 4(1)³ - 1 = 3. Plugging these values into the formula, we get:
X1 = 1 - (-3) / 3
X1 = 2
Now we can repeat the process using X1 as our new X0:
X2 = X1 - f(X1) / f'(X1)
X2 = 2 - (2⁴ - 2 - 3) / (4(2)³ - 1)
X2 ≈ 1.7708
Therefore, the positive solution to the equation x⁴ – x = 3, rounded to four decimal places, is approximately 1.7708.
Step-by-step explanation:
The positive solution to the equation x⁴ – x = 3, estimated using Newton's Method with x₀ = 1 and x₂ as the final estimate, is approximately 1.5329, rounded to four decimal places.
To use Newton's Method to estimate the positive solution to the equation x⁴ – x = 3, we need to find the derivative of the function f(x) = x⁴ – x. This is given by:
f'(x) = 4x³ - 1
We can then use the formula for Newton's Method:
x(n+1) = x(n) - f(x(n)) / f'(x(n))
where x(n) is the nth estimate of the solution.
Starting with x₀ = 1, we can plug this into the formula to get:
x₁ = 1 - (1^4 - 1 - 3) / (4(1^3) - 1) ≈ 1.75
We can then repeat this process using x₁ as the new estimate, to get:
x₂ = 1.75 - (1.75^4 - 1.75 - 3) / (4(1.75^3) - 1) ≈ 1.5329
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Use the figure below to determine the value of the variable and the
lengths of the requested segments. Your answers may be exact or
rounded to the nearest hundredth. The figure may not be to scale.
Using tangents theorem, we can find the value of the missing length,
n = 18.7units.
Define a tangent?"To touch" is how the word "tangent" is defined. The same idea is conveyed by the Latin word "tangere". A tangent, in general, is a line that, while never entering the circle, precisely touches it at one point on its circumference. A circle has a number of tangents. They make a straight angle with the radius.
Here in the diagram,
We can see that as per the central angle and tangent theorem,
AB/BC = ED/DC
⇒ 17/10 = n/11
Cross multiplying:
⇒ 17 × 11 = n × 10
⇒ 10n = 187
⇒ n = 187/10
⇒ n = 18.7
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Now that you have chosen your mode of transportation, use your choice to answer the questions that follow.
What would the cost of your transportation be if you drove:
a. 10 miles? b. 25 miles? c. 42 miles? d. 68 miles?
Make sure to list your chosen mode of transportation and then answer all parts and show your work
(a) The cost of City Bus for driving 10 miles = $3.
(b) The cost of City Bus for driving 25 miles = $7.5.
(c) The cost of City Bus for driving 42 miles = $12.6.
(d) The cost of City Bus for driving 68 miles = $20.4.
We previously choose City Bus as our mode transport since the per mile cost for City Bus is less.
Let the model for City Bus be f(x) = cx + d, where f(x) is total cost and x is number of miles.
From the table of Taxi we get, f(2) = 0.60; f(4) = 1.20; f(6) = 1.80 and f(8) = 2.40.
So, 2a + b = 0.60 and 4a + b = 1.20
(4a + b) - (2a + b) = 1.20 - 0.60
2a = 0.60
a = 0.60/2 = 0.30
Now, f(8) = 2.40
8*0.30 + b = 2.40
2.40 + b = 2.40
b = 2.40 - 2.40 = 0
So the function rule for City Bus is, f(x) = 0.3x.
(a) Total cost to drive 10 miles is,
f(10) = 0.3*10 = 3
(b) Total cost to drive 25 miles is,
f(25) = 0.3*25 = 7.5
(c) Total cost to drive 42 miles is,
f(42) = 0.3*42 = 12.6
(d) Total cost to drive 68 miles is,
f(68) = 0.3*68 = 20.4
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help please ill give brainliest
In the given circle, measure of angle m is 44° and the measure of angle n is 39°. Thus, the value of m is 44 and the value of n is 39
Circle Geometry: Calculating the values of m and nFrom the question, we are to determine the values of m and n in the given circle
From one of the circle theorems, we have that
The angles at the circumference subtended by the same arc are equal. That is, angles in the same segment are equal.
In the given diagram,
Angle m is in the same segment as the angle that measures 44°
Since angles in the same segment are equal,
Measure of angle m = 44°
Also,
Angle n is in the same segment as the angle that measures 39°
Since angles in the same segment are equal,
Measure of angle n = 39°
Hence,
m ∠m = 44°
m ∠n = 39°
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Please help with the 2nd one
Answer:
Step-by-step explanation:
1807
At Kennedy High School, the probability of a student playing in the band is 0. 15. The probability of a student playing in the band and playing on the football team is 0. 3. Given that a student at Kennedy plays in the band, what is the probability that they play on the football team?
The probability that a student at Kennedy High School plays on the football team given that they already play in the band is 2/1 or simply 2.
To solve this problem, we can use conditional probability. We want to find the probability that a student plays on the football team given that they already play in the band.
Let's use the formula for conditional probability:
P(Football | Band) = P(Football and Band) / P(Band)
We know that P(Band) = 0.15, and P(Football and Band) = 0.3.
So,
P(Football | Band) = 0.3 / 0.15
Simplifying, we get:
P(Football | Band) = 2
Therefore, the probability that a student at Kennedy High School plays on the football team given that they already play in the band is 2/1 or simply 2.
Note: This answer may seem unusual because probabilities are typically expressed as fractions or decimals between 0 and 1. However, in this case, we can interpret the result as saying that students who play in the band are twice as likely to also play on the football team compared to the overall population of students.
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(n+3)!/(n+1)! please help immediately
Answer:
(n + 3)(n + 2) or n² + 5n + 6------------------
The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n:
n! = n × (n - 2) × (n - 3) × ... × 2 × 1As per above mentioned definition we see that:
(n + 3)! = (n + 3) × (n + 2) × (n + 1)!Hence the quotient of (n + 3)! and (n + 1)! is:
(n + 3)(n + 2) or n² + 5n + 612
Find the first and second derivatives. S = 15 + 344 - 1 15 S' = S'' =
The first derivative of S is S' = 1/15.
The second derivative of S is S'' = 0.
To find the first derivative (S'):
Starting with the given equation S = 15 + 344 - 1 15, we can simplify it to S = 344 + 15.
We can take the derivative of each term separately since they are added together.
The derivative of a constant (15 and 344) is always 0, so we only need to take the derivative of 1/15.
S' = d/dx (344 + 15)
= d/dx (359)
= 0 + 0 + (d/dx (1/15))
= 1/15
Therefore, the first derivative of S is S' = 1/15.
To find the second derivative (S''):
We need to take the derivative of the first derivative (S').
Since the derivative of a constant is always 0,
we only need to take the derivative of 1/15.
S'' = d/dx (S')
= d/dx (1/15)
= 0
Therefore, the second derivative of S is S'' = 0.
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Ten sixth-grade students reported the hours of sleep they get on nights
before a school day. Their responses are recorded in the dot plot. Looking
at the dot plot, Lin estimated the mean number of hours of sleep to be 8. 5
hours. Noah's estimate was 7. 5 hours. Diego's estimate was 6. 5
hours. Which estimate do you think is best? Solve for the mean to figure out
who was closer. *
Lin
ООО
Noah
Diego
Noah's estimate of 7.5 hours is closest to the actual mean number of hours of sleep reported by the ten sixth-grade students.
We have,
Compare the estimates given by Lin, Noah, and Diego.
Lin's estimate is 8.5 hours.
Noah's estimate is 7.5 hours.
Diego's estimate is 6.5 hours.
The average of these three estimates:
(8.5 + 7.5 + 6.5) / 3
= 22.5 / 3
= 7.5
It appears that Noah's estimate of 7.5 hours is the closest.
Therefore,
Noah's estimate of 7.5 hours is closest to the actual mean number of hours of sleep reported by the ten sixth-grade students.
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The complete question:
Which estimate of the mean number of hours of sleep reported by the ten sixth-grade students is closest to the actual mean: Lin's estimate of 8.5 hours, Noah's estimate of 7.5 hours, or Diego's estimate of 6.5 hours?
HELP!! Which statements describe the end behavior of f(x)?f(x)?
Select two answers.
Answer:
As x approaches ∞, y approaches -∞
As x approaches -2.5, y approaches ∞
Step-by-step explanation:
How many degrees are in the acute angle formed by the hands of a clock at 3:30?
The acute angle formed by the hands of a clock at 3:30 is 75 degrees. An acute angle is an angle that measures less than 90 degrees, and in this case, the hour hand is pointing at the 3, which is a 90-degree angle from the 12, while the minute hand is pointing at the 6, which is a 180-degree angle from the 12.
Find the number of degrees in the acute angle formed by the hands of a clock at 3:30, follow these steps:
Determine the position of the hour hand. At 3:30, the hour hand is halfway between 3 and 4, so it's at 3.5 hours. Convert this to degrees by multiplying by 30 (since there are 360 degrees in a circle and 12 hours on a clock, each hour represents 30 degrees).
So, the hour hand is at 3.5 x 30 = 105 degrees.
Determine the position of the minute hand. At 3:30, the minute hand is on 6, which is 180 degrees around the clock.
Find the difference between the two positions.
Subtract the smaller angle from the larger angle: 180 - 105 = 75 degrees.
The acute angle formed by the hands of a clock at 3:30 is 75 degrees.
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Consider the initial value problem for function y, y (0) = 4. y" + y' - 2 y = 0, y(0) = -5, Find the Laplace Transform of the solution, Y(5) = 4 [y(t)] Y(s) = M Note: You do not need to solve for y(t)
The Laplace transform of the solution to the initial value problem y'' + y' - 2y = 0, y(0) = -5, is Y(s) = (5s + 4) / (s² + s - 2), and Y(5) = 29 / 28.
To find the Laplace transform of the solution to the initial value problem y'' + y' - 2y = 0, y(0) = -5, we can apply the Laplace transform to both sides of the differential equation and use the initial condition to solve for the Laplace transform of y.
Taking the Laplace transform of both sides of the differential equation, using the linearity and derivative properties of the Laplace transform, we get:
L{y'' + y' - 2y} = L{0}
s² Y(s) - s y(0) - y'(0) + s Y(s) - y(0) - 2 Y(s) = 0
s² Y(s) - 5s + s Y(s) + 4 + 2 Y(s) = 0
Simplifying and solving for Y(s), we get:
Y(s) = (5s + 4) / (s²+ s - 2)
To find Y(5), we substitute s = 5 into the expression for Y(s):
Y(5) = (5(5) + 4) / ((5)² + 5 - 2)
Y(5) = 29 / 28
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The table displays data collected, in meters, from a track meet.
14
3
624
What is the median of the data collected?
2 19
332
2.5
02
4.5
03
€
To find the median of the data, we first need to arrange the numbers in order from smallest to largest:
3, 14, 624
Since we have an odd number of data points, the median is the middle number. In this case, the median is 14.
Therefore, the median of the data collected is 14 meters.
The distance between san antonio and houston is 190 miles. nicholas and rose each drove 2/5 of the total distance. if charlie drove the rest of the distance, how many miles did charlie drive?
Charlie drove 90 miles between San Antonio and Houston.
Nicholas and Rose each drove 2/5 of the total distance (190 miles). To find the distance they drove together, multiply 190 miles by 2/5 twice (once for each person):
190 x (2/5) = 76 miles (Nicholas)
190 x (2/5) = 76 miles (Rose)
Together, Nicholas and Rose drove 76 + 76 = 152 miles. To find the remaining distance Charlie drove, subtract this combined distance from the total distance:
190 miles (total) - 152 miles (Nicholas and Rose) = 38 miles (Charlie).
Charlie drove 90 miles between San Antonio and Houston, as Nicholas and Rose each drove 2/5 of the total 190-mile distance, resulting in 152 miles combined, leaving 38 miles for Charlie to cover.
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please show all work so i can better understand. Thank you!
- 2. Find all values of x where f '(x) = 0 for f(x) = arcsin(e2x – 2x).
The only value of x where f'(x) = 0 is x = 0.
Let's find all values of x where the derivative of f(x) = [tex]arcsin(e^(2x) – 2x)[/tex] is equal to 0.
Step 1: Find the derivative f'(x) using the chain rule.
For this, we'll need to differentiate [tex]arcsin(u)[/tex] with respect to u, which is [tex](1/√(1-u^2))[/tex], and then multiply by the derivative of u [tex](e^(2x) – 2x)[/tex]with respect to x. So, f'(x) = [tex](1/√(1-(e^(2x) – 2x)^2)) * d(e^(2x) – 2x)/dx[/tex]
Step 2: Find the derivative of e^(2x) – 2x with respect to x. Using the chain rule and the derivative of [tex]e^u: d(e^(2x) – 2x)/dx = 2e^(2x) – 2[/tex]
Step 3: Combine the derivatives. f'(x) =[tex](1/√(1-(e^(2x) – 2x)^2)) * (2e^(2x) – 2)[/tex]
Step 4: Set f'(x) equal to 0 and solve for x. [tex](1/√(1-(e^(2x) – 2x)^2)) * (2e^(2x) – 2) = 0[/tex]
Since the first part of the product [tex](1/√(1-(e^(2x) – 2x)^2))[/tex] is never 0, we can focus on the second part: [tex]2e^(2x) – 2 = 0[/tex]
Step 5: Solve for x. [tex]2e^(2x) = 2 e^(2x) = 1[/tex]
The only way this is true is when 2x = 0, since [tex]e^0 = 1: 2x = 0 x = 0[/tex]
So, the only value of x where f'(x) = 0 is x = 0.
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A survey was taken by students in 6th, 7th, and 8th grade to determine how many first cousins they have. The results are shown in the box plots below. Use these box plots to answer the questions.
After simplifying, which expressions are equivalent? select three options. (3.4a – 1.7b) (2.5a – 3.9b) (2.5a 1.6b) (3.4a 4b) (–3.9b a) (–1.7b 4.9a) –0.4b (6b – 5.9a) 5.9a – 5.6b
(3.4a – 1.7b) and (–1.7b 4.9a) and (5.9a – 5.6b) are equivalent expressions.
Which expressions are equivalent after simplification?
The question presents a list of six expressions. We need to select three expressions that are equivalent after simplifying them.
One of the expressions is (3.4a - 1.7b), and another is (-1.7b + 4.9a). These two expressions can be simplified to 2.4a - 1.7b.
Another expression is (2.5a - 3.9b), and another is (-3.9b + a). These two expressions can be simplified to 3.5a - 3.9b.
The third expression is 5.9a - 5.6b, which cannot be simplified further.
The three expressions that are equivalent after simplifying are (3.4a - 1.7b) and (-1.7b + 4.9a), (2.5a - 3.9b) and (-3.9b + a), and 5.9a - 5.6b.
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