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3 years ago

pleaseeee explain !!!

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

A is the answer because if we want the unit rate we do

2/3÷2/3 and 15 1/5, we can turn 15 1/5 into a improper fraction first so 76/5 and we do 76/5 ÷2/3 which is equal to

76/5×3/2=114/5

114/5 simplifies into 24 4/5.

so Naomi can travel 24 4/5 gal per mile.

Hope this helps!

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4 divided by the sum of h and 7

Gemiola [76]

Answer:

4/h+7

Step-by-step explanation:

4 is on top of H+7

3 0

3 years ago

Find the area of the triangle

jok3333 [9.3K]

Answer:

Step-by-step explanation:

Area = 25 cm^2

1. Pretend it is a square

2. Multiply the two dimensions (5x10)

3. Take that answer and divide it by two (to get rid of the invisible square) (50/2)

4. Add the measurement

7 0

3 years ago

Read 2 more answers

Plz I need help !!!!!

Ad libitum [116K]

So how you do scientific notation is, however many places you move the decimal point to the left is the number that you put in the power. For example, Mercury's mass is 330,000,000,000,000,000,000,000. You move the decimal point 23 places to the left to get 3.3 x 10^23. ( ^ means power. The little number that is next to the 10)

Now according to the question, the mass of Jupiter is 5.8 x 10^3 multiplied by 3.3 x 10^23 (5,800 x 330,000,000,000,000,000,000,000).

The answer to part A is,
1,941,000,000,000,000,000,000,000,000

Then, 3.2 x 10^2(X)=5.8 x 10^3 x 3.3 x 10^23

I don't know if you know how to do algebra, but this is very simple to solve. All you do is divide bolth sides of the equation by 3.2 x 10^2 leaving X (Earths mass) on the left.

For the right side need to divide 320 by 1,941,000,000,000,000,000,000,000,000. This will be the answer to part B

Hopefully this was helpful enough for you to show your work. Good luck : )

3 0

3 years ago

Given f(x)=1/4(5-x)2 what is the value of f(11)

iren [92.7K]

Answer:

f(x)=1/4(5-x)²

Step-by-step explanation:

f(11)=1/4(5-11)²

1/4(-6)²

1/4(36)

7 0

3 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

￼pleaseeee explain !!!

pleaseeee explain !!!