Ask question

Ask question

All categories

2 years ago

15

Can someone help me on this and explain what most of what to do

1 answer:

irina1246 [14]2 years ago

7 0

For number 11 think of them as decimals and1/2 is .5 and on 12 put them all in as decimals and figure it out and on 13 divide 1/4 and it will get you .25 and add the 6 and the rest all the same rules apply and you just have to figure out which one it is

You might be interested in

When h has the value 4, calculate 16-3h

azamat

h=4

calculate 16-3h

16-(3*4 )

16-12 =4

I hope that's help !

3 0

3 years ago

PLEASE HELP! I am having trouble with both questions

schepotkina [342]

Answer:

someone had the same exact question i just helped him on it sub 4 for x and 1 for h

Step-by-step explanation:

6 0

2 years ago

Write the following numbers in order of size

Anon25 [30]

7/10,.72,75%,4/5,.9 hope this helped

7 0

3 years ago

Read 2 more answers

In the diagram which of the following segments is congruent to segment AB

Valentin [98]

The answer is segment cd

6 0

3 years ago

The sum of 30 times (1/3)^(n-1) from 1 to infinity

sergeinik [125]

Let

$S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}$

Then

$\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}$

and

$S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}$

and as $n\to\infty$ , we end up with

$\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32$

So we have

$\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45$

8 0

3 years ago