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MariettaO [177]

3 years ago

6

Electrical energy (E) is measured in kilowatt hours (kwh) using the formula E = P · t, where P is the power in kilowatts and t i

s the time in hours.
If 104 kwh of electrical energy is created over a period of 13 hours, how many kilowatts of power are used each hour

1 answer:

sveticcg [70]3 years ago

7 0

Kilowatts of power are used each hour
104/13=8

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(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

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2 years ago

A university is comparing the grade point averages of theater majors with the grade point averages of history majors. The univer

AveGali [126]

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2 years ago

Convert 6 7/20<br> into a decimal

Veseljchak [2.6K]

Answer:

6.35

Explanation:

7 divided by 20 is 0.35+6= 6.35

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2 years ago

What is the radius of a circle whose equation is x2 y2 8x – 6y 21 = 0?

andrew11 [14]

we know that

the standard form of the equation of the circle is

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +8x-6y+21=0$

<u>Convert to standard form</u>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+8x) +(y^{2}-6y)=-21$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+8x+16) +(y^{2}-6y+9)=-21+16+9$

$(x^{2}+8x+16) +(y^{2}-6y+9)=4$

Rewrite as perfect squares

$(x+4)^{2} +(y-3)^{2}=4$

$(x+4)^{2} +(y-3)^{2}=2^{2}$

the center of the circle is $(-4,3)$

the radius of the circle is $2\ units$

therefore

<u>the answer is</u>

the radius of the circle is $2\ units$

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3 years ago

Read 2 more answers

Plot the x-intercept of the function f(x) = (x + 4)<br><br> Show Me please!! Need help

Len [333]

Answer : the x-intercept is -4

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3 years ago