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vredina [299]
3 years ago
10

Which of the following are perfect squares? Check all that apply

Mathematics
1 answer:
Kobotan [32]3 years ago
5 0

Answer:

A, C and F

Step-by-step explanation:

A

81 = <em><u>9</u></em><em><u>^</u></em><em><u>2</u></em><em><u> </u></em><em><u> </u></em>

C

100 = <em><u>1</u></em><em><u>0</u></em><em><u>^</u></em><em><u>2</u></em><em><u> </u></em><em><u> </u></em>

F

1 = <em><u>1</u></em><em><u>^</u></em><em><u>2</u></em>

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Select all the rectangles that have the same area, but are not congruent.
san4es73 [151]

Answer:

I don't know the answer if you know then tell me i also need the answer

4 0
3 years ago
Examine this system of linear equations. y – x = 2, x + y = 4 Which is a solution of the system of equations? (1, 3) (2, 2) (3,
Blizzard [7]

Answer:

(1, 3)

Step-by-step explanation:

Solve the following system:

{y - x = 2 | (equation 1)

{x + y = 4 | (equation 2)

Add equation 1 to equation 2:

{-x + y = 2 | (equation 1)

{0 x+2 y = 6 | (equation 2)

Divide equation 2 by 2:

{-x + y = 2 | (equation 1)

{0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{-x+0 y = -1 | (equation 1)

{0 x+y = 3 | (equation 2)

Multiply equation 1 by -1:

{x+0 y = 1 | (equation 1)

{0 x+y = 3 | (equation 2)

Collect results:

Answer: x = 1, y = 3

4 0
3 years ago
Read 2 more answers
Can someone help me.
gtnhenbr [62]

Answer:

\large\boxed{x=\pm\sqrt{10}}

Step-by-step explanation:

x^2=10\Rightarrow x=\pm\sqrt{10}\\\\\text{because}\ (-\sqrt{10})^2=10\ \text{and}\ (\sqrt{10})^2=10.\\\\\text{used}\ (\sqrt{a})^2=a

5 0
4 years ago
Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.
Veseljchak [2.6K]

This equation is separable, as

\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx

Integrate both sides; on the left, expand the fraction as

\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)

Then

\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C

\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C

Since y(0)=1, we get

\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1

so that the particular solution is

\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}

4 0
3 years ago
Please help me i need to complete this by tomorrow
lisov135 [29]

Answer:

Step-by-step explanation: Need to work out your scale factor which is 1.5.

8 * 1.5 = 12

CE = 12

12 * 9 = 108/2 = 59 that is the area of triangle ACE

8 * 6 = 48/2 = 24 that is the area of triangle BCD

ACE - BCD = ABDE

59 - 24 = 35cm^2

8 0
2 years ago
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