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3 years ago

Plz help me with Q:13 And Q:14

Mathematics

2 answers:

Art [367]3 years ago

7 0

For number 13: You have to count how many wedges of cheese there are that are less than 1/2 pound. So there are 5 slices of cheese that is 3/8 pound and 2 slices of cheese that is 1/4 pound. 5+2=7. The answer is B) 7. For number 14: You have to subtract 25 from 193 to get 168 because she made a payment of $25. Next you have to divide 168 by 24, so it equals 7 because she pays $24 each week.The answer is A) 7.

PolarNik [594]3 years ago

5 0

For question 13, you need to look at all the numbers that are less than 1/2 and then count the number of marks in total .
Answer: 7 B
For question 14, you need to first make an equation that suits the circumstances and the situations in this problem.
$25+$24x=$193
$24x=$168
x=7 weeks
x stands for the number of weeks that she has to wait for in order to save enough money to purchase the camera.
Answer: 7 weeks A

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Find the distance between the points (-7,-6) and (5,-6) .

PilotLPTM [1.2K]

Answer:

12 units

Step-by-step explanation:

We are given these two points, (-7,-6) and (5,-6), and we are to find the distance between them.

We know the formula for finding the distance between two points:

Distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So putting in the values of the given coordinates in the above formula to get:

Distance = $\sqrt{(5-(-7))^2+(-6-(-6))^2}$ = $\sqrt{144 + 0} =12$

Therefore, the distance between the two given points is 12 units.

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3 years ago

Identify the slope and y-intercept. y=−12x+7

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3 years ago

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A map has a scale of 1 in. : 5 mi. The distance on the map between two cities is 11.5 inches. Find the actual distance between t

Vladimir [108]

Answer:

57.5 mi

Step-by-step explanation:

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2 years ago

Which expression is equivalent to n + n - .18n

wariber [46]

Answer:7500

Step-by-step explanation:

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2 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