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3 years ago

Please help me I need it please answer it correctly please help me please

Mathematics

1 answer:

jeka57 [31]3 years ago

7 0

On the 7th, Andi had $48 in her account. We want to find when she had $92 less than what she had on the 7th.

First, let's find out how much $92 less is:

48 - 92 = -44

We are looking for when Andi had $-44 in her account. Looking at the number line, the day Andi had $-44 in her account was on the 14th.

The correct answer is C. The 14th.

Hope this helps!! :)

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2:46

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3 years ago

Roberta makes $9 an hour plus a 10.5% commission selling jewelry an hour shift in which she sells $880 worth of jewelry. How muc

timama [110]

Answer:

Roberta earns $164.4 on an 8 hr shift.

Step-by-step explanation:

In one hour Roberta makes $9

So if she works for 8 hours she will be making

=> ( 9 X 8)$

=>72$

Also she earns a commission of 10.5% on selling a jewellery worth $880

So 10.5% of 880 is

=> $\frac{10.5}{100}\times 880$

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Hence Roberta will be earning a total amount of

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=>$164.4

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3 years ago

Derrick made a goal for 2 points during the basketball game. The path of the ball's height in feet can be modeled by y=-x^2-2x+2

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2 years ago

you randomly choose a shirt off the shelf. there is 4 blue shirts 2 black shirts and 2 green shirts and 1 white shirt and 1 red

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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...

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3 years ago