If you want to put a border around a rectangular room what steps would you use to find the length of border needed

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]

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.

[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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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.

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

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

(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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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.

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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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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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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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

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.

Answer:

  A)  x < 15

Step-by-step explanation:

You want the solution to x/3 < 5.

The steps to solving an inequality are basically identical to the steps for solving an equation. There are a couple of differences:

If 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.

Note 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.)

The value of angle x = 114°.

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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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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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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Using tangents theorem, we can find the value of the missing length,

n = 18.7units.

"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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(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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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

From 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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Answer:

Step-by-step explanation:

1807

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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Answer:

------------------

The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n:

As per above mentioned definition we see that:

Hence the quotient of (n + 3)! and (n + 1)! is:

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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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?

Answer:

As x approaches ∞, y approaches -∞

As x approaches -2.5, y approaches ∞

Step-by-step explanation:

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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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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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.

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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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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(3.4a – 1.7b) and (–1.7b 4.9a) and (5.9a – 5.6b) are equivalent expressions.

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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