All categories

2 years ago

PLEASE SHOW WORK FOR BRAINLIEST!!

Mathematics

1 answer:

sergejj [24]2 years ago

6 0

The picture would change the distance from the bridge, the ground, and the jumper.

You might be interested in

You are having a discussion about sequences with your classmate. She insists that the sequence 2, 3, 5, 8, 12 must be either ari

Marizza181 [45]

If arithmetic, then there is a common difference
if geometric, then there is a common ratio

to find the common ratio or difference

if you subtract successive terms from each other, they should have the same result if it is aritmetic
if you divide successive terms by each other, they should have the same result if it is geometric

2,3 and 5

3-2=1
5-3=2
2≠1
not aritmetic

3/2=1.5
5/3=1.66666
1.5≠1.6666
not geometric

she is wrong

4 0

2 years ago

PLEASE ANSWER ASAP! REALLY IMPORTANT! MUST BE CORRECT! POINTS GIVEN!

VMariaS [17]

- 5/12 + ( - 1/4) =

Adding two negatives results in a negative

-1/4 = -3/12

-3/12 + -5/12 = -8/12

-8/12 + 3/12 < Subtract and take the sign of the BIGGER number.

8 - 3 = 5

- 5/12

Answer: - 5/12

8 0

2 years ago

Read 2 more answers

In each of the following pairs of numbers, how much larger is the first number than the second?

lapo4ka [179]

Answer: a. 154 is larger than 27 by 127.

b. 25 is larger than 12 by 13.

c.135 is larger than 127 by 8.

d.46 is larger than 24 by 22.

Step-by-step explanation:

Subtract second number from the first number.

a. 154 and 27

Difference= $154-27=127$

Hence, 154 is larger than 27 by 127.

b. 25 and 12

Difference= $25-12=13$

Hence, 25 is larger than 12 by 13.

c. 135 and 127

Difference= $135-127=8$

Hence, 135 is larger than 127 by 8.

d. 46 and 24

Difference= $46-24=22$

Hence, 46 is larger than 24 by 22.

4 0

2 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

Need help with this

netineya [11]

The answer to your question is C.

3 0

2 years ago