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3 years ago

What is the answer to this questions it is worth 20 points

Mathematics

1 answer:

irinina [24]3 years ago

3 0

<h2>Answer : </h2>

By observing the table we can conclude that for the number of years to travel back double the electricity is required .

i.e. for 5 years = 10 megawatts

therefore, the relationship between x and y can be depicted by the following equation :

$\boxed{ \boxed{y = 2x}}$

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Three angles of quadrilateral ABCD have measures 66, 95 and 114. what is the value of x?

vampirchik [111]

To find the value of x, assuming it is the value of the missing angle in the quadrilateral, you need to know that the sum of the angles in any quadrilateral equals 360.

360 - (66 + 95 + 114) = 85 degrees.

x = 85 degrees

8 0

3 years ago

A cone has a height of 4 in and a radius of 6 in. What is the volume of the cone?

MatroZZZ [7]

Answer:

48π in³

Step-by-step explanation:

$V=\frac{1}{3} \pi r^2h$

Plug in the radius 6 in and the height 4 in

$V=\frac{1}{3} \pi (6)^2(4)\\V=\frac{1}{3} \pi (36)(4)\\V=\frac{1}{3} \pi (144)\\V=48\pi$

Therefore, the volume of the cone is 48π in³.

I hope this helps!

3 0

2 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

a 15-ft ladder is leaning against one wall of an alley 9 ft wide. the ladder slips, its top sliding down the wall, its foot slid

blagie [28]

2x dx/dt +2y dy/dt = 0
x dx/dt + y dy/dt = 0 
9*4 + 12 dy/dt = 0 
dy/dt = - 3 ft/sec

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3 years ago

−1/3 −(−1/2) Brainliest for the right answer

lawyer [7]

The answer would be -1/6 or -0.166667

4 0

3 years ago