Ok i have two algebra questions
1.) subtract: (7x^2-x-+3)
a.) 6x^3+7x^2-x-5
b.) -6x^3+7x^2-x+1
c.) -x^3-x-5
d.) x^2-x+1
2.) f^2*f^4
a.) (2f)^8
b.) (2f)^6
c.) f^8
d.) f^6
explain how you got the last one.
i would appreciate it.
and i will you if its easy to understand
: )

Problem A: (7x^2-x-2)-(-6x^3+3)

You can consider the subtraction similiar to a -1(-6x^3+3). The 1 is implied. So to distribute that negative, we multiply this out.

-6x^3+3 becomes 6x^3 - 3

Our problem is now: 7x^2-x-2 + 6x^3 - 3

Now we add together like items. Let's rearrange for our sanity.

+ 6x^3 + 7x^2 - x - 2 - 3

Simplify: 6x^3 + 7x^2 - x - 5

Unfortionately, outside of factoring or something along those lines, that's as simple as we can get it. So the answer is A.

Problem B: f^2*f^4

Math rules tell us when we multiply the same bases with exponents, we add the exponents together. But the base doesn't change.

4+2 = 6. So: f^6 (D)

That's the easy  Math rules say so.

To expand on that specific rule, consider x^2 * x^3. What at the root are these saying? Multiply 2 copies of X, and then multiply 3 copies of x.

We can rewrite this same equation as x*x * x*x*x. 

Remember that since all of these are multiplication, there is no order of operations that needs to be followed.

If we were simplifying that version of the equation, we would write x ^ 5.

Answer

1 number

explanation

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

 

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

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