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balandron [24]
3 years ago
9

A portion of Raul's check register is shown. His checking account had a balance of $539.50 on April 2. Based on the information

in the check register, what was the balance of Raul's checking account after the transaction on April 13 in dollars and cents?
Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:425.25

Step-by-step explanation:

539.50-

35.50-

23.75-

55.00=

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Sam is a school leader. She wants to decide whether makeup should be allowed in school or not. She
anastassius [24]

Answer:

Following are the sample list can be attached as follows:

Step-by-step explanation:

In the given question some information is missing that is attachment of list which can be attached as follows:

please find the attachment list:

by evaluating the list it will give the answer that the both the samples most want to wear mascara the least want to wear lipstick.

8 0
4 years ago
Mr. Hartman, the school librarian, is ordering new keyboards and mice for all of the school's computer labs. Each keyboard costs
mario62 [17]

Answer:

3 * 20s

Step-by-step explanation:

In order to find out how much Mr. Hartman will spend in total we first need to multiply the price of the keyboard and mouse by the number of computers in a single computer station. Once we have these products we add them together. Finally, we multiply this new value by 3 since there are a total of 3 computer stations. If we turn this into an expression it would be the following...

3 * (13.50s + 6.50s)

We can even simplify this by first adding the cost of the keyboard and mouse and then multiplying by s

3 * 20s

6 0
3 years ago
6 lb 4 oz − 5 lb 10 oz =
FinnZ [79.3K]
The answer to your problem is 10 oz
5 0
3 years ago
Read 2 more answers
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
Lines CD and CE are tangent to circle a. If m∠DAE = 130°, what is the measure of ∠DCE? Tangent CD intersects with circle A at po
inna [77]

The answer is 50º

Step-by-step explanation:

It is the correct and only plausable answer.

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