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3 years ago

Gwen used elimination with multiplication to solve the system:

1 answer:

8 0

The correct answer for this question is this one:

<span>Gwen used elimination with multiplication to solve the system:
2x + 6y = 3
x − 3y = −5

Her work to find x is shown. Complete the explanation of her error. Then solve the system. If necessary, enter your answer as a reduced fraction.
2(x − 3y) = −5 --> 2(x-3y) = 2(-5)
2x − 6y = −5 --> 2x - 6y = -10
2x + 6y = 3 --> 2x + 6y = 3
2x − 6y = −5 --> 2x - 6y = -10
4x + 0y = −2 --> 4x = -7
x = −2/4 --> <u>x = -7/4</u>

x - 3y = -5
4(-7/4 - 3y = -5)
-7 - 12y = -20
-12y = -13
<u>y = 12/13</u>

Gwen forgot to multiply the right side by 2.
What is the solution as an ordered pair</span>

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$A(4)=-4\pi$

Calculation of A(8):

we can plug x=8

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$A(8)=-4\pi +\frac{1}{2}\times 4\times 4$

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Calculation of A(12):

we can plug x=12

$A(12)=\int\limits^12_0 f{x} \, dx$

we can break into two parts

$A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx$

now, we can find area and then combine them

$A(12)=-4\pi +\frac{1}{2}\times 8\times 4$

$A(12)=-4\pi +16$

Calculation of A(14):

we can plug x=14

$A(14)=\int\limits^14_0 f{x} \, dx$

we can break into two parts

$A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx$

now, we can find area and then combine them

$A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2$

$A(14)=-4\pi +16-1$

$A(14)=-4\pi +15$

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Read 2 more answers

The volume of a sphere whose diameter is 18 centimeters is _ cubic centimeters. If it’s diameter we’re reduced by half, it’s vol

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<h2>Answer:</h2>

<u>First Part</u>

Given that

$Volume = \frac{4}{3} \pi r^{3}$

We have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{Diameter}{2})^{3} = \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}$

<u>Second Part</u>

Given that

$Volume = \frac{4}{3} \pi r^{3}$

If the Diameter were reduced by half we have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}$

This shows that the volume would be $\frac{1}{8}$ of its original volume

<h2>Step-by-step explanation:</h2>

<u>First Part</u>

Gather Information

$Diameter = 18cm$

$Volume = \frac{4}{3} \pi r^{3}$

Calculate Radius from Diameter

$Radius = \frac{Diameter}{2} = \frac{18}{2} = 9$

Use the Radius on the Volume formula

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi 9^{3}$

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

$Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3$

We can see now that one of the 3s can be already divided by the 3 in the denominator

$Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3$

Finally, since we can't simplify anymore we just calculate it's volume

$Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}$

$Volume \approx 3053.63 cm^{3}$

<u>Second Part</u>

Understanding how the Diameter reduced by half would change the Radius

$Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}$

Understanding how the Radius now changes the Volume

$Volume = \frac{4}{3}\pi r^{3}$

With the original Diameter, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}\\\\$

If the Diameter were reduced by half, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}$

But we can see that the numerator is exactly the original Volume!

This shows us that the Volume would be $\frac{1}{8}$ of the original Volume if the Diameter were reduced by half.

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2 years ago