Ask question

Ask question

All categories

4 years ago

13

I don’t understand how to do it

1 answer:

Troyanec [42]4 years ago

5 0

Answer:

I believe the answer is 0, because almost any equation multiplied by 0 will be 0.

You might be interested in

PLEASE ANSWER ASAP! please

FinnZ [79.3K]

Answer:

3 ÷ 0.5

Step-by-step explanation:

Since each large square is 1 whole, and theirs five out of 10 columns that are distinctivly a different orange, it'd be 3 divided by 0.5.

5 0

3 years ago

Pls help me answer this question i need it asap. This is a good offer :)

tensa zangetsu [6.8K]

<u>Answer:</u>

The product is 21/32.

<u>Step-by-step explanation:</u>

<u>Write the product:</u>

=> 2 5/8 x 2/8

<u>Convert it to improper fraction:</u>

=> 21/8 x 2/8

<u>Multiply:</u>

=> 42/64

<u>Simplify:</u>

=> 21/32

Hence, <u>the product is 21/32.</u>

Hoped this helped.

$BrainiacUser1357$

7 0

3 years ago

Read 2 more answers

A container can hold no more than 80 lb. Rick fills the box with books that each weigh 2 lb and a statue that weighs 10 lb. Whic

Bad White [126]

2B + 10 ≤ 80.....................

6 0

3 years ago

PLEASE HELP ME BEST ANSWER GETS BRAINLIEST NO SCAMS PLEASE!

fomenos

Answer:

False

Step-by-step explanation:

People reading in a library are most likely going to like to read books

8 0

3 years ago

Read 2 more answers

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago