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3 years ago

Find the coordinates of E' after a dilation with a scale factor of 2.

Mathematics

1 answer:

V125BC [204]3 years ago

4 0

Answer:

(more information is needed, the origin of the dilation and the coordinates for E)

You might be interested in

What is the answers for expanded form 72.3 7.023 0.723

kvasek [131]

Answer:

look below

Step-by-step explanation:

72.3 = 70 + 2 + .3

7.023 = 7 + .02 + .003

.723 = .7 + .02 + .003

Hopefully this helps you :)

pls mark brainlest ;)

7 0

4 years ago

Find the equivalent fraction or decimal for each number. 1 over 4

pentagon [3]

2/8, 2 Over 8 because all your doing is multiplying 1x2 and 4x2

7 0

4 years ago

What else equals 42 and 30 for multiply

aev [14]

Answer:

1 times 42 and 6 times 7 for 42

1 times 0 and 5 times 6

(I did this question in school once)

7 0

3 years ago

In one week, 390 people answered a telephone survey.

QveST [7]

Answer:

150 people

Step-by-step explanation:

First, look over the chart and find the rows that have listed as 26 - 30 and 31 - 35.

Find out how high the row goes by looking at the number on the side of the gram.

26 - 30 goes to 60

31 - 35 goes to 90

Then add the 2 numbers together and you will find your total

60 + 90 = 150 people

8 0

3 years ago

The spherical balloon is inflated at the rate of 10 m³/sec. Find the rate at which the surface area is increasing when the radiu

rjkz [21]

The balloon has a volume $V$ dependent on its radius $r$ :

$V(r)=\dfrac43\pi r^3$

Differentiating with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$

If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

$10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}$

The surface area of the balloon is

$S(r)=4\pi r^2$

and differentiating gives

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}$

so that at the moment the radius is 3 m, its area is increasing at a rate of

$\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}$

4 0

3 years ago