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3 years ago

If a point is equidistant from the endpoints of a segment then so

1 answer:

6 0

Answer: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.

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Destiny works on commission as an office supply saleswoman. She sold a computer that cost $700 and received a $126 commission. W

Lisa [10]

To figure out the answer to this problem, you must first ask yourself this:

How much is 126 of 700
 or
What is 126 divded by 700

The answer is 0.18.

So, that means that 0.18 or 18 percent might be your answer.

Now, lets try 18 percent of 700, to see if we get 126.

18% times 700

The anwer is 126.

SO:

$126 is 18 percent of 700.

18 percent of her sale is Destiny's commission.

7 0

3 years ago

The admission at the movies is $5 for children and $10 for adults. The total amount collected was

sesenic [268]

Answer:

Number of children, x= 400

Number of adults, y = 100

Step-by-step explanation:

Let the number of children be x

Let the number of adults be y

Given : Total number of people, x + y = 500 --------(1)

Total amount collected, 5x + 10y = 3000 ---------(2)

(1) => y = 500 - x

Substitute y in (2) : 5x + 10(500 - x) = 3000

5x + 5000 - 10x = 3000

-5x = 3000 - 5000

-5x = -2000

x = 400

Substitute x in (1) : 400 + y = 500

y = 500 - 400

y = 100

4 0

3 years ago

What is the answer in fraction form to 1/2 + 2/3

gtnhenbr [62]

Answer:1.17

Step-by-step explanation:

8 0

3 years ago

Please help! Correct Answers only!

mafiozo [28]

Answer:

$12000

Step-by-step explanation:

Using the given formula :

I = P × R × T

I = 40000 × 15/100 × 2

I = $12000

5 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