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4 years ago

8

What is 5 (x-4) + 3x - 9x + 7?

1 answer:

Westkost [7]4 years ago

3 0

Answer:

-x - 13

Step-by-step explanation:

5(x-4) + 3x - 9x + 7

First we use the distributive property to remove the parenthesis

5x -20 + 3x - 9x + 7

then we combine the like terms

-x - 13

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Which is bigger 4 kilometers or 4400 meters?

Musya8 [376]

1km=1000m
4400m=(4400/1000)km
4400m=4.4km
therefore, 4400meters is bigger

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3 years ago

Help me and I will crown you please!!!

Lina20 [59]

All you do is divide 96 divided by 6. That is your equation. Then when you divide it, you find that he will put 16 on each shelf. All of the shelves will have the same number of books. The next one all you do is add all of them up and it will equal 43. So he has 43 dvd in total. Now if he wants to only keep eight of the dvds you subtract eight from 43. That equals 35. Now all you do is divide that by 5 to see how many he will give each friend. Once you're finished with that you find that he will give 7 to each friend.
So, the answers to each problem are 16 with the equation using distributive property 2(44 + 2) divided by 6 and 7.
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6 0

4 years ago

Read 2 more answers

Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

<u>case B)</u> $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

<u>case C)</u> $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

<u>case A)</u> $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

<u>the answer is</u>

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

6 0

3 years ago

Read 2 more answers

Marie plays the state lottery. During the lotteries in June, she lost $10, won $20, lost $8, lost $12, and won $25. What was the

xenn [34]

The answer to your question is Marie won 15$

7 0

3 years ago

Read 2 more answers

A jar contains jelly beans that are red, yellow, blue, and purple. Several people reach in and grab handfuls of jelly beans, the

dem82 [27]

Total jellybeans pulled out: 15 + 20 + 12 + 8 = 55

Proportion of yellow jellybeans: 20 / 55 = 4/11

(4/11) * 22 = 8.

Answer: 8

3 0

4 years ago