Find the cost for 812 kwhr of electricity at 7.3 cents per kwhr.

Mathematics

1 answer:

Over [174]3 years ago

5 0

812*0.073= the cost would be 59.276$. and I'm good with math if u need further help

Hope this helps

You might be interested in

What are solutions to the equation below x^2 +3x-18=0

Ivenika [448]

$x^2 +3x-18=0\\ x^2-3x+6x-18=0\\ x(x-3)+6(x-3)=0\\ (x+6)(x-3)=0\\ x=-6 \vee x=3$

5 0

3 years ago

Read 2 more answers

What is 67/100 as a decimal

ASHA 777 [7]

<span>67/100 as a decimal.

Divide numerator by denominator. Answer = decimal form.

67/100
67/100 = 0.67

67/100 as a decimal = 0.67
</span>

4 0

3 years ago

Read 2 more answers

If a hypothesis test leads to the rejection of the null hypothesis, a _____.

Butoxors [25]

Answer:

Step-by-step explanation:

Hope it helped u

6 0

2 years ago

Trey designs a new game for cell phones. The game has 10 levels.

Vilka [71]

10 * 38 = 380 seconds

7 0

3 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago