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4 years ago

Hi can you help me please?)

Mathematics

1 answer:

Ad libitum [116K]4 years ago

8 0

<em>To solve for a variable, you just need to isolate the variable to one side.</em>

For this, just divide both sides by r and <u>your answer will be $\frac{d}{r}=t$ </u>

For this, divide both sides by nR and <u>your answer will be $\frac{PV}{nR}=T$ </u>

Firstly, multiply both sides by T: $AT=FV-OV$

Next, subtract FV on both sides of the equation: $AT-FV=-OV$

Lastly, multiply both sides by -1, and <u>your answer will be $-AT+FV=OV$ </u>

Firstly, multiply both sides by 1000: $1000C=Wtc$

Lastly, divide both sides by tc and <u>your answer will be $\frac{1000C}{tc}=W$ </u>

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How many boxes of markers must be sold for the company’s income to equal the cost of production

Margaret [11]

C. 200

They earn $2.50 per box of markers, meaning that to make $500 they have to sell 200. [500/2.5=200]

6 0

2 years ago

Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3$

I'm going to again insert some zero placeholders to help me line up my like terms for the addition.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^6+0x^4+x^3+0x^2+2x+3$

$0x^6-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3+0x^2+0x+0$

----------------------------------------------------Adding the three products!

$x^6+x^4+4x^3-2x^2-x+3$

8 0

4 years ago

This is index btw people

Setler [38]

A.) 4^2

B.) 5^4

C.) 3^1 x 7^4

D.) 2^2 x 9^4

Hope this helps you!!!!!!

3 0

3 years ago

PLEASE SOLVE. Students are selling candy bars. They earn $0.50 for each one they sell. They want to earn at least $455. Write an

kari74 [83]

So each candy bar is .50 and they want to reach 455.00

we see how many .50's can go into 455.00

we notice that .50 times 2=1
so if we multiply the 455 1's by 2 we get how many candy bars to sell or
910.00

6 0

3 years ago

Read 2 more answers

Find AC please, help is very much appreciated.

fenix001 [56]

Answer:

AC ≈ 67.9

Step-by-step explanation:

15² = 9 (3x + 1 + 9)

225 = 9 (3x + 10)

225 = 27x + 90

135 = 27x

x ≈ 19.29

AC = (3x + 1 + 9)

AC = 3 (19.2857...) + 1 + 9

AC = 57.857... + 1 + 9

AC = 57.857... + 10

AC ≈ 67.9

4 0

3 years ago