3 years ago

(6+8k^2+8k^3)+(k^2+4k-6)

Mathematics

1 answer:

lianna [129]3 years ago

3 0

$8 {k}^{3} + 9 {k}^{2} + 4k$

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Expand 3(n+6) This for the test

MrRa [10]

Step-by-step explanation:

expanded = 3n + 18

plz mark my answer as brainlist plzzzz if you find it useful ☺️.

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2 years ago

What is the common denominator of 1/a+1/b in the complex fraction ((1/a-1/b)/(1/a+1/b)? A^2b^2 a-b a+b ab

S_A_V [24]

Answer:

D. ab

Step-by-step explanation:

1/a + 1/b =
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3 years ago

What line is parallel y=5x-3

Korvikt [17]

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3 years ago

Thomas is sketching the graph of a quadratic function. he is asking his friends about how many x-intercepts his function can hav

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Answer:

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3 years ago

Read 2 more answers

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

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2 years ago