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2 years ago

What is the product? (-2x -9y2) (-4x-3)

Mathematics

1 answer:

lutik1710 [3]2 years ago

6 0

Answer:

8x^4+36x^3y^2

Step-by-step explanation:

(-2x-9y^2)(-4x^3)

8x^4+36x^3y^2

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Simplify the equation -2(p + 4)2 - 3 + 5p. What is the simplified expression in standard form? ASAP PLEASE!! ILL GIVE EXTRA POIN

Katena32 [7]

So you distrubute and get

-2p-8+2-3+5p
add like terms
3p-9
ANSWER: 3p-9

4 0

2 years ago

Read 2 more answers

Eight more than 3 times a number

Sladkaya [172]

Answer:

3x + 8

Step-by-step explanation:

7 0

3 years ago

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403, 892 in expanded form

eimsori [14]

Hey there!

<h3>"Expanded form" is basically, setting it up in a addition form to give you that result</h3>

$4 = 400,000 \\ 0 = 0 \\ 3 = 3,000 \\ 8 = 800 \\ 9 = 90 \\ 2 =2 \\ \\ \\ \\ Answer: 400,000 + 0 + 3,000 + 800 + 90 + 2$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

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2 years ago

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system. [Start 2 By 3 Matrix 1s

Leni [432]

Answer: h = 9

Step-by-step explanation: A system of linear equations is consistent when it has at least one solution.

The matrix given is:

$\left[\begin{array}{ccc}-15&21&h\\5&-7&-3\end{array}\right]$

Transform this matrix in a row-echelon form:

$\left[\begin{array}{ccc}-15&21&h\\5&-7&-3\end{array}\right]$ $R_{2} = 3R_{2}+R_{1}$ $\left[\begin{array}{ccc}-15&21&h\\0&0&-9+h\end{array}\right]$

For this row-echelon form to have solutions:

-9 + h = 0

h = 9

For this system to be consistent: h = 9.

3 0

3 years ago

This is really confusing me:<br> Simplify so that your answers contain only positive exponents.

xxTIMURxx [149]

Answer:

see below

Step-by-step explanation:

For simplifying expressions of this sort, there are four rules of exponents that come into play;

$a^ba^c=a^{b+c}\\\\(ab)^c=a^cb^c\\\\(a^b)^c=a^{bc}\\\\\dfrac{1}{a^b}=a^{-b}\ \quad\text{which means}\quad a^b=\dfrac{1}{a^{-b}}$

I find it convenient to eliminate the fractions by adding the exponents, then rewrite any negative exponents as denominator factors.

$23. \quad\dfrac{1}{x^{-2}}=x^2\\\\24. \quad\dfrac{mn}{m^2n^3}=m^{1-2}n^{1-3}=m^{-1}n^{-2}=\dfrac{1}{mn^2}\\\\25. \quad\dfrac{k^{-2}}{k^{-3}}=k^{-2-(-3)}=k\\\\26. \quad\left(\dfrac{m^2}{n^3}\right)^3=\dfrac{m^{2\cdot 3}}{n^{3\cdot 3}}=\dfrac{m^6}{n^9}\\\\27. \quad\dfrac{x^2y^3z^4}{x^{-3}y^{-4}z^{-5}}=x^{2-(-3)}y^{3-(-4)}z^{4-(-5)}=x^5y^7z^9$

3 0

3 years ago