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3 years ago

Write the product in simplest form: -8w * (-w)

Mathematics

1 answer:

Hitman42 [59]3 years ago

7 0

8w

-8w times -w is like -8*-1 which would just make the number positive.

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HELP PLEASE!!! I don’t understand these math questions at all and it’s due in 2 hours, I’m freaking out

nataly862011 [7]

Answer:

1. The top and bottom lines (-5,3) (8,3) is parallel to line (-2,-2) (11,-2). The sides Lines (-5,3) (-2,-2) is parallel to line (8,3) (11,-2).

2. None of the line create a right angle. None are able to create a box at the angle

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3 years ago

It will cost Amber an estimated $3,723 a year in tuition to attend her preferred college. She has been putting $350 a month in a

kenny6666 [7]

Answer:

Yes; she will have $4008 for other expenses.

Step-by-step explanation:

First you find out how much she saved by multiplying 350 by 36 as she saved for 3 years which is 36 months.

$350 \times 36 = 12600$

After that you need to work out the amount she needs in total for her tuition by multiplying 4 by 3723 which is the amount needed per year.

$3723 \times 4 = 14892$

If her parents add to her account half the amount she saved that means that they give her an additional 6300 as that's half of 12600.

This means she has 18900 and if you take away 14892 you will be left with 4008 for expenses.

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3 years ago

What is the original price:$65.00; Markdown: 12%

alexandr1967 [171]

The answer to this question is $57.02

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3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

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

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

3 years ago

Help plz I really need the help

Colt1911 [192]

Answer:

the answer might be either one of the four

3 0

3 years ago