Ask question

Ask question

All categories

3 years ago

10

Allie collected 15 more cans of food than Peyton Ling collected 8 more than Ali ling collected 72 cans of food how many cans of

food did Peyton collect

2 answers:

a_sh-v [17]3 years ago

8 0

Peyton collected 90
Allie collected 105
Ali collected 72

will u mark brainliest please

Mars2501 [29]3 years ago

4 0

Your answer is ninety five

You might be interested in

Without building the graph find the coordinates of the point of intersection of the lines gIven by the equations y=2x+3 and 3x+y

Natali5045456 [20]

Answer:

intersection is (1,5)

Step-by-step explanation:

plug one equations into the other

y=2x+3

3x+y=8

3x+2x+3=8

simplify

5x+3=8

subtract 3

5x=5

x=1

plug x into one equation

y=2+3=5

intersection is (1,5)

7 0

3 years ago

17 feet in 4 minutes<br><br>pls hurry

IrinaK [193]

Answer:

4.25

Step-by-step explanation:

17/4

= 4.25

I hope it helps! Have a great day!

Muffin ^^

8 0

2 years ago

Read 2 more answers

Help please! I have no idea how to do this

Ivanshal [37]

Answer:

it is greater than

Step-by-step explanation:

4 0

3 years ago

Find the average value of the function g(x) = <img src="https://tex.z-dn.net/?f=%5Cfrac%7Blnx%7D%7Bx%7D" id="TexFormula1" title=

aleksley [76]

Answer:

$\displaystyle average=\frac{1}{2(e-1)}$

Step-by-step explanation:

<u>Average Value of a Function</u>

Given a function g(x), we can compute the average value of g in a given interval (a,b) with the equation:

$\displaystyle average=\frac{1}{b-a}\int_{a}^{b} g(x)dx$

We use the given data

$\displaystyle average=\frac{1}{e-1}\int_{1}^{e} \frac{lnx}{x}dx$

We now compute the indefinite integral with a u-substitution

$\displaystyle I=\int \frac{lnx}{x}dx$

We'll use the substitution u=lnx, du=dx/x. Then

$\displaystyle I=\int u.du$

Integrating

$\displaystyle I=\frac{u^2}{2}$

Since u=lnx

$\displaystyle I=\frac{ln^2x}{2}$

The average value is

$\displaystyle average=\frac{1}{e-1}\left|\frac{ln^2x}{2} \right|_1^e$

$\displaystyle average=\frac{1}{e-1}\left(\frac{ln^2e}{2}-\frac{ln^21}{2} \right )$

Since lne=1, and ln1=0

$\displaystyle average=\frac{1}{e-1}\left(\frac{1}{2}-0 \right )$

$\displaystyle average=\frac{1}{2(e-1)}$

3 0

4 years ago

Use the figure to answer the question.

zavuch27 [327]

Answer:

I don't see a figure sorry

8 0

3 years ago

Read 2 more answers