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3 years ago

3+x) 5 Solve it for me, and tell me how to do it EXPLANATION NEEDED

1 answer:

3 0

I’m not too sure of what the equation is but I made a guess let me know if this was the right equation please I can redo it.

But, when solving this equation you are trying to find the value of x. To do that you have to get rid of the 3.

You would either add or subtract the 3 on both sides.

When the three is gone you should have somthing like

x = 5
- 3

Which you would move on to subtract the 5-3 and get 2.

x= 2

Here’s a picture of how I solved it

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1,284 to the nearest 100 place

prisoha [69]

1,300 because 8 is over 5 and it rounds the 100th place ,(which is 2), to become a 3 which makes the answer 1,300.

7 0

3 years ago

Read 2 more answers

100 POINT QUESTION

Valentin [98]

its not 100 points it 8 but i will still help you

the answer is 7.2 i believe

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3 years ago

(4^p*4^1)^3 simplifies to 4^15 what does p equal?

zzz [600]

When an exponent is outside the parentheses, you "distribute" it to the other exponents inside the parentheses.

(4^p)^3 would become 4^3p

(4^1)^3 would become 4^3

When the bases are the same, you just simply add the exponents.

Let's just ignore the bases so it'll look less confusing.

3p + 3 = 15

Subtract 3 from both sides.

3p = 12

Divide both sides by 3.

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3 years ago

A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair a

vesna_86 [32]

$Chair=x$

$Table=y$

$Desk=z$

$\begin{Bmatrix}x+y&=&40\\y+z&=&83\\x+z&=&77\end{matrix}$

keep the first row as normal, then in the other ones, we can isolate Y and X

$\begin{Bmatrix}x+y&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace at first row...

$\begin{Bmatrix}(77-z)+(83-z)&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}160-2z&=&40\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&40-160\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}-2z&=&-120\\y&=&83-z\\x&=&77-z\end{matrix}$

$\begin{Bmatrix}z&=&60\\y&=&83-z\\x&=&77-z\end{matrix}$

now we can replace the Z to discovery the other value

$\begin{Bmatrix}z&=&60\\y&=&83-60\\x&=&77-60\end{matrix}$

$\boxed{\boxed{\boxed{\begin{Bmatrix}Chair&=&17\\Table&=&23\\Desk&=&60\end{matrix}}}}$

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3 years ago

Consider the following linear programming problem.

givi [52]

Answer:

This is not worth only 5 pts have a nicer heart. -_-

Step-by-step explanation:

4 0

2 years ago