Ask question

Ask question

All categories

PSYCHO15rus [73]

3 years ago

6

Clancy and Ethan are starting new books so far clancey has read 1/4 of his book which has 240 total pages and ethan has read 2/5

of his book which has 170 total pages who has read more pages so far clancey or eathan

1 answer:

Lemur [1.5K]3 years ago

6 0

Answer:

Ethan(68)

Step-by-step explanation:

clanxey=1\4*240=60

eathan=2\5*170=68

You might be interested in

Describe the slope of the line. (-3,1) and (2,-2) The slope is _____, find the slope m=?

devlian [24]

Answer:

$m = - \frac{3}{5}$

Step-by-step explanation:

4 0

3 years ago

Solve the eqaution using the quadratic formula 3x^2=2(2x+1)

steposvetlana [31]

Answer:

the answer is x=1,−13

Step-by-step explanation:

.

8 0

3 years ago

Read 2 more answers

. Which of the following fractions are equivalent? A. 3 4 and 6 7 B. 2 3 and 12 13 C. 1 9 and 4 12 D. 1 6 and 3 18

Ainat [17]

Answer:

A-3/4 is the only equivalent fraction.

Step-by-step explanation:

The rest of the fractions that you have listed does not pull up as equivalent.

Could you go to my page and help me figure out how to work out my sheet for me? I need help ASAP! Someone has already tried to help me, but they did not make sense to part of my sheet and they have not yet responded. Would you be willing to help? I will mark you brainliest if correct, and give you +45 points! Thank you!

3 0

2 years ago

I need help with #10-12 Plzzz thank you guys sm

frutty [35]

Here is the answer buddy
124.32

5 0

3 years ago

14. 2057 Q.No. 1(a) Sum to infinity:<br>1 + 3x + 5x2 + 7x3 +... (-1<x<1).

aleksandrvk [35]

The sum appears to be

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n$

I'll assume you want to find out what function this sum converges to.

Let

$f(x)=\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

with -1 < <em>x</em> < 1. Differentiating gives

$f'(x)=\dfrac1{(1-x)^2}=\displaystyle\sum_{n=0}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty(n+1)x^n$

So we have

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n=f'(x)+xf'(x)$

$\displaystyle\sum_{n=0}^\infty(2n+1)x^n=\frac{1+x}{(1-x)^2}$

5 0

3 years ago