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2 years ago

SHOW WORK! PLEASE HELP ASAP IF YOU CAN :(

Mathematics

1 answer:

Sedbober [7]2 years ago

4 0

Answer:

so what you have to do first is simplify into variables

S=A 3/4 stack of index cards

C= 3/160

first you can do this with a calculator, divide 3/4 of an inch by 3/160 of an inch.

3/4÷3/160=40

Here is my work shown :)

Male the division into multiplication and swap the numbers in 3/160 to 160/3

3/4 x 160/3

For fraction multiplication, multiply the numerators and then multiply the demoninators to get

3 x 160/4 x 3= 480/12

Divide 480 by 12 which is forty

there you go :) sorry if the answers are messed up and there are spelling errors. but in conclusion the answer is 40

By the way, awesome pfp, I just started watching Toilet Bound Hanako earlier tonight and it's very good so far

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RSB [31]

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2 years ago

* if my savings of $x grows 10% each year how much money would i have in 1 year

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2 years ago

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irina [24]

Option D:

$\left(y^{2}+3 y+7\right)\left(8 y^{2}+y+1\right)=8 y^{4}+25 y^{3}+60 y^{2}+10y+7$

Solution:

Given expression is $\left(y^{2}+3 y+7\right)\left(8 y^{2}+y+1\right)$ .

<u>To find the product of the expression:</u>

$\left(y^{2}+3 y+7\right)\left(8 y^{2}+y+1\right)$

Multiply each term of the first term with each term of the 2nd term.

$=y^{2}\left(8 y^{2}+y+1\right) +3 y\left(8 y^{2}+y+1\right) +7\left(8 y^{2}+y+1\right)$

Using the exponent rule: $a^m \cdot a^n = a^{m+n}$

$=\left(8 y^{4}+y^3+y^2\right) +\left(24 y^{3}+3y^2+3y\right) +\left(56 y^{2}+7y+7\right)$

$=8 y^{4}+y^3+y^2+24 y^{3}+3y^2+3y+56 y^{2}+7y+7$

Arrange the terms with same power.

$=8 y^{4}+y^3+24 y^{3}+y^2+3y^2+56 y^{2}+7y+3y+7$

$=8 y^{4}+25 y^{3}+60 y^{2}+10y+7$

Hence option D is the correct answer.

$\left(y^{2}+3 y+7\right)\left(8 y^{2}+y+1\right)=8 y^{4}+25 y^{3}+60 y^{2}+10y+7$

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3 years ago

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Answer:

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2 years ago

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Uppose a triangle has a,b and c what must be true?

alexdok [17]

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Right: b^2 + c^2 = a^2

Acute: b^2 + c^2 < a^2

A^2 + B^2 + c^2 - 2bc*cosA
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2 years ago

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