Factor the polynomial:

Mathematics

1 answer:

den301095 [7]2 years ago

3 0

$a^8-a^2b^6=a^2(a^6-b^6)\qquad\text{use}\ (x^n)^m=x^{nm}\\\\=a^2[(a^3)^2-(b^3)^2]\qquad\text{use}\ (x-y)(x+y)=x^2-y^2\\\\=a^2(a^3-b^3)(a^3+b^3)\qquad\text{use}\ x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)\\\\=\boxed{a^2(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)}$

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Find the dot product of the position vectors whose terminal points are (14, 9) and (3, 6).

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Answer:

Step-by-step explanation:

The formula for the dot product of vectors is

u·v = |u||v|cosθ

where |u| and |v| are the magnitudes (lengths) of the vectors. The formula for that is the same as Pythagorean's Theorem.

$|u|=\sqrt{14^2+9^2}$ which is $\sqrt{277}$

$|v|=\sqrt{3^2+6^2}$ which is $\sqrt{45}$

I am assuming by looking at the above that you can determine where the numbers under the square root signs came from. It's pretty apparent.

We also need the angle, which of course has its own formula.

$cos\theta=\frac{uv}{|u||v|}$ where uv has ITS own formula:

uv = (14 * 3) + (9 * 6) which is taking the numbers in the i positions in the first set of parenthesis and adding their product to the product of the numbers in the j positions.

uv = 96.

To get the denominator, multiply the lengths of the vectors together. Then take the inverse cosine of the whole mess:

$cos^{-1}\theta=\frac{96}{111.64676}$ which returns an angle measure of 30.7. Plugging that all into the dot product formula: