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3 years ago

Anyone donate help?

Mathematics

1 answer:

sweet [91]3 years ago

3 0

Answer:

4(2x-y) (2x+y)

Step-by-step explanation:

The area of the picture is 2y*2y = 4y^2

The area of the picture and the frame is 4x*4x = 16x^2

The area of the frame is The area of the picture and the frame minus The area of the picture

The area of the frame = 16x^2- 4y^2

Factor out a 4

4(4x^2 -y^2)

This is the difference of squares

a^2 – b^2 = (a + b)(a – b)

a^2 = 4x^2 so a=2x

b^2 = y^2 so b=y

4(4x^2 -y^2) = 4(2x-y) (2x+y)

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What must be added to both sides of the equation x – 12 = 15 to isolate the variable?

pishuonlain [190]

Answer:

c. 12

Step-by-step explanation:

To isolate the variable, you need to cancel out whatever's on the same side of the variable by using the inverse of the operation.

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2 years ago

For the dilation with center (0,0) shown on the graph, your friend says the scale factor is 8/3

olchik [2.2K]

Answer:

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2 years ago

Joelle is working on a school report in a word processor. She wants to place an oversized image which is 30.51 \text{ cm}30.51 c

ehidna [41]

Answer:

the horizontal position for the left edge of the image is -4.46 cm.

Step-by-step explanation:

The horizontal midpoint of the page is:

21.59 / 2 = 10.79 cm

The horizontal midpoint of the image is given as:

30.51 / 2 = 15.25 cm

These midpoints must be on the same point in the axis.

By taking the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795 cm.

The distance that goes beyond this leftmost edge is computed as;

This is on the negative side on the axis.

Thus the horizontal position for the left edge of the image is -4.46 cm.

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2 years ago

PLEASE HELP!!!!!! ASAP

deff fn [24]

the first one. The camera focuses on the audience

6 0

3 years ago

A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

Klio2033 [76]

Answer:

k = 0.032
P0 = 131.836

Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

Comparing this to the desired form, we see that ...

$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$

So, the approximate equation for P is ...

$P(t)=131.836\cdote^{0.032t}$

And the parameters of interest are ...

k = 0.032
P0 = 131.836

4 0

2 years ago