Answer:
A factorial is the product of all positive integers from 1 to a given number. In this case, the given number is 8, so the factorial would be:
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
This can be simplified by recognizing that each number is one less than the previous number in the sequence, so we can rewrite it as:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
where the exclamation mark denotes a factorial. Thus, the given expression can be written as 8!
the picture below is being enlarged by a scale factor of 2.5. how many inches of farming will the picture require? base 5 in high4 in
A. 12.5 in
B. 20 in
C. 45 in
D. 65 in
The inches of farming the picture require is 45 inches
How many inches of farming will the picture require?From the question, we have the following parameters that can be used in our computation:
Base = 5 in
High = 4 in
Scale factor = 2.5
The inches of farming the picture require is calculated as
Perimeter = 2 * (Base + High) * Scale factor
Substitute the known values in the above equation, so, we have the following representation
Perimeter = 2 * (5 + 4) * 2.5
Evaluate
Perimeter = 45
Hence, the inches of farming the picture require 45 inches
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A city's population in the year x=1953 was y=2,695,750. In 1971 the population was 2,694,850. Compute a slope of the population growth or decline and choose the most accurate statement
The negative slope indicates a decline in population over the 18-year period. The most accurate statement based on this information is that the city's population experienced a decline of approximately 50 people per year on average between 1953 and 1971.
To compute the slope of the population growth or decline, we need to use the formula:
slope = (y2 - y1) / (x2 - x1)
where y2 is the final population, y1 is the initial population, x2 is the final year, and x1 is the initial year.
Plugging in the values we have:
slope = (2,694,850 - 2,695,750) / (1971 - 1953)
slope = -900 / 18
slope = -50
The negative slope indicates a decline in population over the 18-year period.
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Given m||n, find the value of x
The value of x is 32.
A set of angles that are between the two line parallels but on each side of the transverse are referred to as alternate external or exterior angles. The measure of the alternate external or exterior angles is equal.
According to the alternate outside angle theorem, alternate exterior angles are regarded as congruent angles or degrees of equal magnitude when two parallel lines intersected by a transversal.
Given that the lines m and n are parallel, and the angles are [tex](3x-2)^{0}[/tex] and [tex](2x+30)^{0}[/tex].
The given angles are alternate external or exterior angles. So, they must be equal.
[tex](3x-2)^{0} = (2x+30)^{0}[/tex]
Rearrange the above equation to find the value of x as follows,
[tex]3x-2x=30+2[/tex]
[tex]x=32[/tex]
Hence, the value of x is 32.
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3.3 Dr Seroto travelled from his office directly to the school 45 km away. He travelled at an average speed of 100 km per hour and arrived at the school at 11:20. Verify, showing ALL calculations, whether Dr Seroto left his office at exactly 10:50. The following formula may be used: Distance = average speed x time
[tex]distance= average sped \times time[/tex]
Answer: We can use the formula Distance = Average speed x time to verify whether Dr Seroto left his office at exactly 10:50.
Let t be the time Dr Seroto left his office. Then, the time he arrived at the school can be expressed as:
t + (Distance/Average speed) = 11:20
We know that the distance is 45 km and the average speed is 100 km/hour. Substituting these values, we get:
t + (45/100) = 11:20
We need to convert the time on the right-hand side to hours. 11:20 can be written as:
11 + 20/60 = 11.33 hours
Substituting this value, we get:
t + 0.45 = 11.33
Solving for t, we get:
t = 11.33 - 0.45
t = 10.88 hours
This is not equal to 10:50, which is 10.83 hours. Therefore, Dr Seroto did not leave his office at exactly 10:50.
Let a be a number. Find the n-vector b for which,bc =p/(a).This means that the derivative of the polynomial at a given point is a linear functionof its coefficients.
We have found the n-vector b for which bc = p/(a), where p is the polynomial with coefficients c0, c1, ..., cn. Let's start by writing out the polynomial:
[tex]c0 + c1x + c2x^2 + ... + cnx^n[/tex]
The derivative of this polynomial with respect to x is:
[tex]c1 + 2c2x + 3c3x^2 + ... + ncnx^(n-1)[/tex]
At the point x=a, the derivative becomes:
[tex]c1 + 2c2a + 3c3a^2 + ... + ncn*a^(n-1)[/tex]
We want this to be a linear function of the coefficients c0, c1, ..., cn. That means we need to find a vector b such that:
[tex]c1 + 2c2a + 3c3a^2 + ... + ncna^(n-1) = b0c0 + b1c1 + ... + bncn[/tex]
Let's compare coefficients of c0, c1, ..., cn on both sides:
c1 = b1c1
2c2a = b2c2
[tex]3c3a^2 = b3c3[/tex]
...
[tex]ncna^(n-1) = bn*cn[/tex]
We can simplify these equations by dividing both sides by ci (assuming ci is not zero):
b1 = 1
b2 = 2a/c2
[tex]b3 = 3a^2/c3[/tex]
...
[tex]bn = n*a^(n-1)/cn[/tex]
So the vector b we're looking for is:
b = [1, 2a/c2, [tex]3a^2[/tex]/c3, ...,[tex]n*a^(n-1)/cn][/tex]
And if we multiply b by the coefficient vector c, we get:
bc = c1 + 2c2a + [tex]3c3a^2[/tex] + ...[tex]+ ncn*a^(n-1)[/tex] = the derivative of the polynomial at x=a
Therefore, we have found the n-vector b for which bc = p/(a), where p is the polynomial with coefficients c0, c1, ..., cn.
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Find area bounded by y = in(x)/x, y = 0, and x = e¹⁹. (Express numbers in exact form. Use symbolic notation and fractions where needed.)
The area bounded by y = in(x)/x, y = 0, and x = e¹⁹ is 361/2 square units.
To find the area bounded by the curves y = ln(x)/x, y = 0, and x = e¹⁹, we need to integrate the function ln(x)/x with respect to x over the interval [1, e¹⁹].
∫[1, e¹⁹] ln(x)/x dx
To solve this integral, we use integration by parts with u = ln(x) and dv/dx = 1/x dx.
∫ ln(x)/x dx = ∫u dv = uv - ∫v du
where v = ln(x) and du/dx = 1/x dx.
∫ ln(x)/x dx = ln(x)^2/2 |[1, e¹⁹] - ∫[1, e¹⁹] (1/x)(ln(x)/2) dx
Evaluating the definite integral at the limits gives:
ln(e¹⁹)²/2 - ln(1)²/2 = 361/2
So the area bounded by the curves y = ln(x)/x, y = 0, and x = e¹⁹ is 361/2 square units.
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At the Fisher farm, the weights of zucchini squash are Normally distributed. Which standardized weight represents the top 10% of the zucchinis?
Find the z-table here.
–1. 64
–1. 28
1. 28
1. 64
The standardized weight which represents the top 10% of the zucchinis from the z-score for the fisher farm the weights of zucchini squash are Normally distributed is 1.28.
Standardized normal distribution = Z given by ,
Z = (X - μ)/σ
Here, X is sample, is μ mean and is σ standard deviation.
At the Fisher farm, the weights of zucchini squash are Normally distributed.
For the normal distribution,
The value mean be 0 and standard deviation be 1.
Z = (X - μ)/σ
μ = 0 , σ = 1
Z = (X -0)/1
Z = X
For the top 10% of the zucchinis, the value of α is 0.9. From the table for this value the z score is,
Z = X
Z = 1.28
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Write a derivative formula for the function.
f(x) = 2x√7x+8 + 96
The derivative formula of the given function f(x) is:
f'(x) = 2√(7x+8) + 7x/(√(7x+8))
This formula gives the value of the derivative of the function f(x) for any value of x.
How we find derivative?The derivative of f(x) using the product rule and chain rule:
f'(x) = 2√(7x+8) + 2x(1/2)(7x+8)^(-1/2)(7)
f'(x) = 2√(7x+8) + 7x/(√(7x+8))
Simplify the derivative
The derivative of the function f(x) is given by f'(x) = 2√(7x+8) + 7x/(√(7x+8))
In this formula, the first term represents the derivative of the function 2x√(7x+8) using the chain rule, and the second term represents the derivative of the function 96, which is a constant and has a derivative of zero.
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Solve the inequality for x 4x-1 grater then -5
The solution of the inequality 4x-1 grater then -5 is x > -1
Solving the inequality for x
From the question, we have the following parameters that can be used in our computation:
4x-1 grater then -5
Express properly
so, we have the following representation
4x - 1 > -5
Add 1 to both sides of the inequality
so, we have the following representation
4x > -4
Divide both sides by 4
x > -1
Hence, the solution of the inequality is x > -1
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Several scientists decided to travel to South America each year beginning in 2001 and record the number of insect species they encountered on each trip. The table shows the values coding 2001 as 1, 2002 as 2, and so on. Find the model that best fits the data and identify its corresponding R2 value
The best-fitting model for the data and its corresponding R2 value need to be calculated.
How to model data?To find the model that best fits the data and its corresponding R2 value, we would need to perform linear regression analysis on the data. However, since the data table is not provided, we cannot provide an answer to this question.
Linear regression analysis is a statistical method used to model the relationship between two variables. In this case, the variables are the year and the number of insect species encountered on each trip. By analyzing the data, we can determine the equation of the line that best fits the data and the R2 value, which represents the proportion of the variance in the data that is accounted for by the model. A higher R2 value indicates a better fit between the model and the data.
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Joe bought a computer that was 20% off the regular price of $1280. What was the discount Joe received?
The discount Joe received is $256 and Joe paid only $1024 for the computer after availing the 20% discount.
Joe purchased a computer that was priced at $1280. However, he was able to avail a discount of 20% off the regular price. To find out the discount that Joe received, we can use a simple formula.
Discount = Regular Price x Discount Rate
In this case, the regular price is $1280 and the discount rate is 20%. Therefore, the discount Joe received is:
Discount = $1280 x 0.20
Discount = $256
So, Joe received a discount of $256 on his purchase of the computer. This means that he paid only $1024 for the computer after availing the 20% discount. It's always important to calculate discounts before making any purchase to ensure you're getting the best deal possible.
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Deation 15 of 25
mat is the point-slope equation of a line with alope -3 that contains the point
A.y-4--3(x-8)
By+4=3(x-8)
ay+4=3(x+8)
Dy-4=3(x*8)
Answer:
y = -3x + 20.
Step-by-step explanation:
The correct point-slope equation of a line with slope -3 that contains the point (8,-4) is:
y - (-4) = -3(x - 8)
Expanding the right-hand side:
y + 4 = -3x + 24
Subtracting 4 from both sides:
y = -3x + 20
Therefore, the answer is not given in any of the options provided. The correct equation is y = -3x + 20.
Barbara’s Bigtime Bakery baked the world’s largest chocolate cake. (It was also the world’s worstcake, as 343 people got sick after eating it. ) The length was 600 cm, the width 400 cm, and the height 180 cm. Barbara and her two assistants, Boris and Bernie, applied green peppermint frosting on the four sides and the top. How many liters offrosting did they need for this dieter’s nightmare? One liter of green frosting covers about 1200 cm²
The total liters of frosting needed is 500, under the condition that the length was 600 cm, the width 400 cm, and the height 180 cm.
In order to evaluate the amount of frosting needed, we have to evaluate the surface area of the cake. The surface area of the cake is the summation of the areas of all its sides.
Here the area of each side is equivalent to its length multiplied by its width. Then the area of the given top is equivalent to its length multiplied by its width.
Then the evaluated surface area of the cake is
2 × (length × height + width × height) + length × width
= 2 × (600 cm × 180 cm + 400 cm × 180 cm) + 600 cm × 400 cm
= 2 × (108000 cm² + 72000 cm²) + 240000 cm²
= 2 × 180000 cm² + 240000 cm²
= 600000 cm²
Hence, one liter of green frosting covers about 1200 cm².
600000 cm² / 1200 cm² per liter = 500 liters
Therefore, Barbara and her assistants needed 500 liters of frosting for their dieter's nightmare.
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Find the mass and center of mass of a wire in the shape of the helix x = t, y = 2 cos t, z = 2 sin t, 0 ≤ t ≤ 2π, if the density at any point is equal to the square of the distance from the origin.
The center of mass is given by:
(xbar, ybar, zbar) = (Mx/m, My/m, Mz/m)
= (0, 16/(3π), 16/(3π))
To find the mass of the wire, we need to integrate the density function over the length of the wire. The length of the wire can be found using the arc length formula:
ds = sqrt(dx^2 + dy^2 + dz^2)
= sqrt(1 + 4sin^2(t) + 4cos^2(t)) dt
= sqrt(5) dt
Integrating this from 0 to 2π gives us the length of the wire:
L = ∫_0^(2π) sqrt(5) dt
= 2πsqrt(5)
Now we can find the mass of the wire:
m = ∫_0^(2π) ρ ds
= ∫_0^(2π) (x^2 + y^2 + z^2) ds
= ∫_0^(2π) (t^2 + 4cos^2(t) + 4sin^2(t)) sqrt(5) dt
= 4πsqrt(5)
To find the center of mass, we need to find the moments about each coordinate axis:
Mx = ∫_0^(2π) ρ x ds
= ∫_0^(2π) t(t^2 + 4cos^2(t) + 4sin^2(t)) sqrt(5) dt
= 0 (due to symmetry)
My = ∫_0^(2π) ρ y ds
= ∫_0^(2π) 2cos^2(t) (t^2 + 4cos^2(t) + 4sin^2(t)) sqrt(5) dt
= 32π/(3sqrt(5))
Mz = ∫_0^(2π) ρ z ds
= ∫_0^(2π) 2sin^2(t) (t^2 + 4cos^2(t) + 4sin^2(t)) sqrt(5) dt
= 32π/(3sqrt(5))
Finally, the center of mass is given by:
(xbar, ybar, zbar) = (Mx/m, My/m, Mz/m)
= (0, 16/(3π), 16/(3π))
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Let x1 > 1 and xn+1 := 2−1/xn for n ∈ N. Show that xn is bounded and monotone. Find the limit. Prove by induction
We have shown that xn is bounded and monotone increasing, and its limit is √2. First, we will show that xn is bounded and monotone increasing by induction:
Base Case: For n = 1, we have x1 > 1, which is true.
Inductive Hypothesis: Assume that xn > 1 for some n = k and show that xn+1 > xn for n = k.
Inductive Step:
We have xn+1 := 2−1/xn
Since xn > 1, we have 1/xn < 1
Therefore, 2−1/xn > 2−1/1 = 1/2
So, xn+1 > 1/2
Since xn > 1, we have xn+1 = 2−1/xn < 2−1/1/ = 1
So, 1/2 < xn+1 < 1
Therefore, xn is bounded and monotone increasing.
Next, we will find the limit of xn as n → ∞:
Let L = lim xn as n → ∞
Then, taking the limit on both sides of xn+1 = 2−1/xn, we get:
L = 2−1/L
Multiplying both sides by L, we get:
L2 = 2−1
Solving for L, we get:
L = ±√2
Since xn > 1 for all n, we have L > 1. Therefore, L = √2.
Thus, we have shown that xn is bounded and monotone increasing, and its limit is √2.
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Qn in attachment
.
..
Answer:
option d
Step-by-step explanation:
24
pls mrk me brainliest (* ̄(エ) ̄*)
The point (-5,. 7) is located on the terminal arm of ZA in standard position. A) Determine the primary trigonometric ratios for ZA If applicable, make Sure yoU rationalize the denominator: b) Determine the primary trigonometric ratios for _B with the Same sine as ZA; but different signs for the other two primary trigonometric ratios If applicable, make sure you rationalize the denominator: c) Use a calculator to determine the measures of ZA and _B, to the nearest degree:
(a)We can use these values to calculate the primary trigonometric ratios:
sin(ZA) = o/h ≈ 0.139
cos(ZA) = a/h ≈ -0.998
tan(ZA) = o/a ≈ -0.14
(b) The same sine as ZA but different signs for the other two primary trigonometric ratios can be found by reflecting point (-5, 0.7) across the x-axis.
(c)We use inverse trigonometric functions on primary ratios ZA ≈ 7 degrees, B ≈ -7 degrees.
(a)How to calculate primary trigonometric ratios?To determine the primary trigonometric ratios for ZA, we first need to find the values of the adjacent, opposite, and hypotenuse sides of the right triangle that contains point (-5, 0.7) as one of its vertices. We can use the Pythagorean theorem to find the hypotenuse:
h = sqrt((-5)² + 0.7²) ≈ 5.02
The adjacent side is negative since the point is to the left of the origin, so:
a = -5
The opposite side is positive since the point is above the x-axis, so:
o = 0.7
Now we can use these values to calculate the primary trigonometric ratios:
sin(ZA) = o/h ≈ 0.139
cos(ZA) = a/h ≈ -0.998
tan(ZA) = o/a ≈ -0.14
(b) How trigonometric ratios can be found by reflecting point?To find a point B with the same sine as ZA but different signs for the other two primary trigonometric ratios, we can reflect point (-5, 0.7) across the x-axis. This gives us point (-5, -0.7), which has the same sine but opposite sign for the cosine and tangent:
sin(B) = sin(ZA) ≈ 0.139
cos(B) = -cos(ZA) ≈ 0.998
tan(B) = -tan(ZA) ≈ -0.14
(c) How to determine measures of nearest degree?To find the measures of ZA and B to the nearest degree, we can use inverse trigonometric functions on their primary ratios. Using a calculator, we get:
ZA ≈ 7 degrees
B ≈ -7 degrees (Note: this is equivalent to 353 degrees since angles are periodic).
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Solve the equation
Complete using the provide data and solve
Answer:
VM = 20
Step-by-step explanation:
Basic proportionality theorem or Thale's theorem:
If a line is drawn parallel to one side of a triangle to intersect the other sides in two side in distinct points, the other two sides are divided in the same ratio.
VN = VT - NT
= 49 - 14
= 35
[tex]\sf \dfrac{VM}{MU} = \dfrac{VN}{NT}\\\\\\\dfrac{VM}{8}=\dfrac{35}{14}\\\\\\\dfrac{VM}{8}=\dfrac{5}{2} \\\\\\VM = \dfrac{5}{2}*8\\\\VM=5*4\\\\\boxed{\bf VM = 20}[/tex]
Need help with this question
Answer:
11
Step-by-step explanation:
23-12=11
A straight line ax+by=16.it passes through a(2,5) and b(3,7).find values of a and b
The values of a and b that satisfy the equation of the line and pass through points A(2,5) and B(3,7) are a = 3 and b = 2.
To find the values of a and b, we need to use the coordinates of points A and B and the equation of the line ax+by=16.
First, we substitute point A(2,5) into the equation to get:
a(2) + b(5) = 16
Next, we substitute point B(3,7) into the equation to get:
a(3) + b(7) = 16
We now have two equations with two unknowns, which we can solve simultaneously.
Multiplying the first equation by 3 and the second equation by -2, we get:
6a + 15b = 48
-6a - 14b = -32
Adding the two equations, we eliminate the a variable and get:
b = 2
Substituting b = 2 into one of the original equations, we get:
2a + 10 = 16
Solving for a, we get:
a = 3
Therefore, the values of a and b that satisfy the equation of the line and pass through points A(2,5) and B(3,7) are a = 3 and b = 2.
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The quantity of a substance can be modeled by the function Z(t) that satisfies the dᏃ differential equation dZ/dt = 1/20Z. One point on this function is Z(1) = 140. Based on this model, use a linear approximation to the graph of Z at = 1 to estimate the quantity of the substance at t = 1.2
The estimated quantity of the substance at t = 1.2 is approximately 140.018.
The given differential equation is: dZ/dt = 1/20Z
Separating variables and integrating, we have:
∫ Z dZ = ∫ 1/20 dt
1/2 Z^2 = 1/20 t + C
where C is the constant of integration.
Using the given initial condition Z(1) = 140, we can solve for C:
1/2 (140)^2 = 1/20 (1) + C
C = 9800 - 70 = 9730
So, the equation that models the quantity of the substance is:
1/2 Z^2 = 1/20 t + 9730
Now, we can use linear approximation to estimate the quantity of the substance at t = 1.2, based on the information at t = 1.
The linear approximation formula is:
L(x) = f(a) + f'(a) * (x - a)
where a is the known point and f'(a) is the derivative of the function at a.
In this case, a = 1, so we have:
Z(1.2) ≈ Z(1) + Z'(1) * (1.2 - 1)
To find Z'(1), we take the derivative of the function:
Z(t) = √(40t + 194600)
Z'(t) = (40/2) * (40t + 194600)^(-1/2) * 40
Z'(t) = 800/(40t + 194600)^(1/2)
So, at t = 1, we have:
Z'(1) = 800/(40(1) + 194600)^(1/2) ≈ 0.0898
Now, we can use the linear approximation formula to estimate Z(1.2):
Z(1.2) ≈ Z(1) + Z'(1) * (1.2 - 1)
Z(1.2) ≈ 140 + 0.0898 * 0.2
Z(1.2) ≈ 140.018
Therefore, based on this model, the estimated quantity of the substance at t = 1.2 is approximately 140.018.
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An eighth-grade student estimated that she needs $9,500 for tuition and fees for each year of college. She already has $5,000 in a savings account. The table shows the projected future value of the account in five years based on different monthly deposits. Initial Balance (dollars) $5,000 $5,000 $5,000 $5,000 Monthly Deposit (dollars) $100 $200 $300 $400 Account Value in Five Years (dollars) $11,000 $17,000 $23,000 $29,000 Problem The student wants to have enough money saved in four years to pay the tuition and fees for her first two years of college. Based on the table, what is the minimum amount she should deposit in the savings account every month? A $200 B $100 C $300 D $400
$23,000 is more than the $19,000 needed for tuition and fees for the first two years, the minimum amount the student should deposit in the savings account every month is $300 (option C).
What is the account value in five years with a monthly deposit of $300?The minimum monthly deposit the student should make in order to save enough money for her first two years of college tuition and fees in four years, we need to calculate how much she would need to have in her savings account at the end of four years.
Since the student estimated she needs $9,500 per year for tuition and fees, she will need $19,000 for her first two years. Since she already has $5,000 in her savings account, she needs to save an additional $14,000 in four years.
Looking at the table, we can see that the account value in five years with a monthly deposit of $200 is $17,000. To find out how much the account value would be in four years, we need to calculate the future value of $17,000 with a 4-year time frame and an annual interest rate of 0%, which gives:
Future value = $17,000 x (1 + 0%)^(4 x 12/12) = $17,000
Since the account value with a $200 monthly deposit is only $17,000 after 5 years, which is not enough to cover the $19,000 needed for tuition and fees for the first two years, the student needs to make a higher monthly deposit.
Looking at the table again, we can see that the account value in five years with a monthly deposit of $300 is $23,000. To find out how much the account value would be in four years, we need to calculate the future value of $23,000 with a 4-year time frame and an annual interest rate of 0%, which gives:
Future value = $23,000 x (1 + 0%)^(4 x 12/12) = $23,000
$23,000 is more than the $19,000 needed for tuition and fees for the first two years, the minimum amount the student should deposit in the savings account every month is $300 (option C).
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Jill's neighborhood has a mean of 3 children per household.
What happens to the mean if a family with 7 children moves
away?
The mean decreases.
The mean remains the same.
The mean increases.
Can you guys help pls
Copy and complete the table of values for y=x2+x
What numbers replace A B and C?
X -2 -1 0 1 2
Y 2 A B 2 C
The value of A, B and C from the given equation are 0, 0 and 6 respectively.
The given equation is y=x²+x.
Here, the table is
X -2 -1 0 1 2
Y 2 A B 2 C
When x=-1
y=(-1)²-1
y=0
So, A=0
When x=0
y=(0)²+0
y=0
So, B=0
When x=2
y=(2)²+2
y=6
So, C=6
Therefore, the value of A, B and C are 0, 0 and 6 respectively.
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hellp if you can love you
Answer:
Circumference = 56.52 cmStep-by-step explanation:
It's given that, Radius of the circle is 9 cm.
We know that Circumference of the circle is calculated as 2πr
where,
π = 3.14Substituting the required values,
Circumference = 2 × 3.14 × 9
= 6.28 × 9
= 56.52 cm
Hence the required circumference of the circle is 56.52 cm
At what value(s) of x does cos x = 8x? X= (Use a comma to separate answers as needed. Type an integer or decimal rounded to two decimal places as needed.) Use Newton's method to obtain the third approximation, X2, of the positive fourth root of 4 by calculating the third approximation of the right 0 of f(x)= x⁴ - 4. Start with x0 = 1. The third approximation of the fourth root of 4 determined by calculating the third approximation of the right of f(x) = x⁴ - 4, starting with x0 = 1, is (Round to four decimal places.)
The third approximation of the fourth root of 4, starting with x0 = 1, is approximately 1.7321
To find the third approximation, X2, of the positive fourth root of 4 using Newton's method, we will follow these steps:
1. Define the function f(x) = x^4 - 4 and its derivative f'(x) = 4x^3.
2. Start with an initial guess x0 = 1.
3. Apply Newton's method formula to find the next approximation: x1 = x0 - f(x0) / f'(x0).
4. Repeat the process for the second and third approximations.
Step 1:
f(x) = x^4 - 4
f'(x) = 4x^3
Step 2:
x0 = 1
Step 3:
x1 = x0 - f(x0) / f'(x0) = 1 - (1^4 - 4) / (4 * 1^3) = 1 - (-3 / 4) = 1 + 0.75 = 1.75
Step 4:
x2 = x1 - f(x1) / f'(x1) = 1.75 - (1.75^4 - 4) / (4 * 1.75^3) ≈ 1.7321
The third approximation of the fourth root of 4 determined by calculating the third approximation of the right of f(x) = x^4 - 4, starting with x0 = 1, is approximately 1.7321 (rounded to four decimal places).
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F(x) =∣x∣ g(x)=∣x+2∣ we can think of g as a translated (shifted) version of f. complete the description of the transformation.
The transformation of f(x) to g(x) involves a horizontal shift and a reflection. The function g(x) is a transformed version of f(x) obtained by translating f(x) to the left by 2 units along the x-axis.
Specifically, to obtain g(x) from f(x), we first shift f(x) two units to the left, and then we take the absolute value of the result.
This means that the graph of g(x) will be the same as the graph of f(x) for all values of x greater than or equal to -2, but will be reflected across the y-axis for all values of x less than -2.
In other words, the transformation of f(x) to g(x) involves a horizontal shift and a reflection.
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3. (3 points) For ordinary differential equation
X =1- ƛx³6
with ƛ > 0, compute the update Ax= x(t+h) - x(t) using
⚫ Euler's method
⚫ the implicit Euler method
⚫ the midpoint method.
The following are the updates for the given ordinary differential equation using Euler's method, the implicit Euler method, and the midpoint method.
To compute the update Ax for the given ordinary differential equation using Euler's method, we first need to discretize the time domain. Let t0 be the initial time and tn = t0 + nh be the time after n steps of size h. Then, using Euler's method, we have:
xn+1 = xn + hf(xn, tn)
where f(xn, tn) = 1 - ƛxn³/6. Therefore,
Ax = xn+1 - xn = h(1 - ƛxn³/6)
Using the implicit Euler method, we have:
xn+1 = xn + hf(xn+1, tn+1)
where f(xn+1, tn+1) = 1 - ƛxn+1³/6. Solving for xn+1, we get:
xn+1 = (xn + h)/[1 + ƛh/6(xn+1)²]
which is a nonlinear equation that needs to be solved iteratively at each step. Therefore, the update Ax becomes:
Ax = xn+1 - xn
Using the midpoint method, we have:
xn+1 = xn + hf(xn+½h, tn+½h)
where f(xn+½h, tn+½h) = 1 - ƛ(xn+½h)³/6. Therefore,
xn+1 = xn + h(1 - ƛxn³/6 + 3ƛx²n h/4)
and the update Ax becomes:
Ax = xn+1 - xn = h(1 - ƛxn³/6 + 3ƛx²n h/4)
These are the updates for the given ordinary differential equation using Euler's method, the implicit Euler method, and the midpoint method.
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1. Simplify (Write each expression without using the absolute value symbol. )
|120+x| if x<-120
2. Simplify (Write each expression without using the absolute value symbol. )
|x-120| if x<-120
3. Simplify (Write each expression without using the absolute value symbol. )
|x-(-12)| if x>-12
4. Simplify (Write each expression without using the absolute value symbol. )
|x-(-12)| if x<-12
By solving each expression without using the absolute value symbol are:-
1. -x-120 if x<-120
2. -(x-120) if x<-120
3. x+12 if x>-12
4. -(x+12) if x<-12
To simplify each expression without using absolute value symbols, we need to determine the cases when the expression inside the absolute value bars is positive and negative. Then we can remove the absolute value bars and simplify the expression accordingly.
For the first expression, |120+x|, if x is less than -120, then x+120 will be negative. Therefore, we can simplify the expression as -x-120.
For the second expression, |x-120|, if x is less than -120, then x-120 will also be negative. Therefore, we can simplify the expression as -(x-120).
For the third expression, |x-(-12)|, if x is greater than -12, then x-(-12) is positive. Therefore, we can simplify the expression as x+12.
For the fourth expression, |x-(-12)|, if x is less than -12, then x-(-12) is negative. Therefore, we can simplify the expression as -(x+12).
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The table shows the dimensions of four boxes.
Drag tiles to order the volumes of the boxes from least to greatest
The order of the volumes of the boxes from least to greatest is Box D, Box B, Box C, Box A. Therefore, the correct option is D.
To determine the order of the volumes of the boxes from least to greatest, we will first calculate the volume of each box using the formula:
Volume = Length × Width × Height.
Hence,
1. Box A: Volume = 2in × 4.5in × 6in = 54 cubic inches
2. Box B: Volume = 6in × 2.5in × 3in = 45 cubic inches
3. Box C: Volume = 5in × 4.5in × 2.25in = 50.625 cubic inches
4. Box D: Volume = 2.5in × 2.25in × 3in = 16.875 cubic inches
Now, arrange the volumes in ascending order:
Box D (16.875), Box B (45), Box C (50.625), Box A (54)
Thus, the correct answer is D: Box D, Box B, Box C, Box A.
Note: The question is incomplete. The complete question probably is: The table shows the dimensions of four boxes. Which is the order of the volumes of the boxes from least to greatest?
Length Width Height
Box A 2in; 4.5in; 6in
Box B 6in; 2.5in; 3in
Box C 5in; 4.5in; 2.25in
Box D 2.5in; 2.25in; 3in
A) Box A Box B, Box C, Box D B) Box A, Box C, Box B, Box D C) Box B, Box D, Box A, Box C D) Box D, Box B, Box C, Box A.
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