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Ivenika [448]
2 years ago
7

Don’t understand how to get the answer

Mathematics
2 answers:
otez555 [7]2 years ago
8 0

Answer:

Additive inverse is the number that you could add to a number to get 0. What could you add to -14 to get 0?

If you need more help, feel free to just ask in the comments.

pishuonlain [190]2 years ago
5 0

Answer:

The answer is 14 because it's the opposite of -14

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factor 3a^5-18a^3+6a^2, I know the answer is 3a^2(a^3-6a+2) but I am so lost on how to get the factor of 3a^2..How in the world
rusak2 [61]

Answer:

\huge\boxed{3a^2(a^3-6a+2)}

Step-by-step explanation:

\sf 3a^5-18a^3+6a^2\\\\HCF = 3a^2\\\\Take \ 3a^2 \ common\\\\= 3a^2(a^3-6a+2)\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
5 0
2 years ago
What is the product of − 3/11 and − 2/15? please explain.
emmainna [20.7K]

ANSWER

The product is

\frac{2}{55}

EXPLANATION

To find the product of

-  \frac{3}{11}

and

\frac{ - 2}{15}

means we should multiply the two fractions.

We multiply to obtain,

-  \frac{3}{11}  \times  -  \frac{2}{15}

=  -  \frac{3}{11}  \times  -  \frac{2}{3 \times 5}

Cancel the common factors:

=  -  \frac{1}{11}  \times  -  \frac{2}{5}

Now, multiply the numerators separately and the denominators too separately.

=  \frac{ - 1 \times  - 2}{11 \times 5}

This simplifies to,

= \frac{2}{55}

4 0
2 years ago
What is the slope of the line passing through (-5,2) and (1,8)
nikdorinn [45]

Answer:

go to slope calclculator

Step-by-step explanation:

6 0
2 years ago
What is the coefficient in the term k?<br> A.0<br> B.1<br> C.2<br> D.k
Amiraneli [1.4K]

9514 1404 393

Answer:

  1

Step-by-step explanation:

The constant multiplier of k in the term "k" is 1.

The coefficient is 1.

4 0
2 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3
KatRina [158]

Answer:

<h2>B. x = -1, x = -3, y = 0</h2>

Step-by-step explanation:

g(x)=\dfrac{x-2}{x^2+4x+3}\\\\vertical\ asymptote:\\\\x^2+4x+3=0\\x^2+x+3x+3=0\\x(x+1)+3(x+1)=0\\(x+1)(x+3)=0\iff x+1=0\ \vee\ x+3=0\\\\\boxed{x=-1\ \vee\ x=-3}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x-2}{x^2+4x+3}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(\frac{1}{x}-\frac{2}{x^2}\right)}{x^2\left(1+\frac{4}{x}+\frac{3}{x^2}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{3}{x^2}}=\dfrac{0}{1}=0\\\\\boxed{y=0}

4 0
3 years ago
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