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2 years ago

1

Mathematics

1 answer:

nevsk [136]2 years ago

3 0

B. 0

Explanation: at sea level is 0, below sea level would be a negative such as -2, and above sea level is any number above zero, such as 2.

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What is the approximate area of the shaded sector in the circle shown below

Zepler [3.9K]

I believe it is C hope this helps

3 0

3 years ago

Read 2 more answers

Jean sold cupcakes and cookies yesterday. Each cupcake sold for $2.25 and each cookie sold for $0.50. At the end of the day, Jea

slega [8]

Answer:number of cupcakes sold is 13

Number of cookies sold is 27

Step-by-step explanation:

Let x represent the number of cupcakes that Jane sold.

Let y represent the number of cookies that Jane sold.

Each cupcake sold for $2.25 and each cookie sold for $0.50. At the end of the day, Jean had sold $42.75 worth of cookies and cupcakes.

This means that

2.25x + 0.5y = 42.75 - - - - - - - - - -1

she sold 40 cupcakes and cookies combined, it means that

x + y = 40

Substituting x = 40 - y into equation 1, it becomes

2.25(40 - y) + 0.5y = 42.75

90 - 2.25y + 0.5y = 42.75

- 2.25y + 0.5y = 42.75 - 90

- 1.75y = - 47.25

y = - 47.35/-1.75

y = 27

x = 40 - y

x = 40-27 = 13

4 0

3 years ago

500

statuscvo [17]

Answer: The required ratio would be 3 miles : 4 minutes.

Step-by-step explanation:

Since we have given that

Speed of a cheetah = 45 miles per hour

So, it means

Distance = 45 miles

Time = 1 hour = 60 minutes

So, the ratio of the distance it travels to the time it takes would be

Distance : Time

$45\ miles:60\ minutes\\\\=3\ miles :4\ minutes$

Hence, the required ratio would be 3 miles : 4 minutes.

3 0

2 years ago

Benny has saved three thousand nine hundred cents over four days from selling lemonade.How many dollars does Benny have?

saw5 [17]

30 Dollars.

Hope this helps please mark this as Brainliest answer,

6 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