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oh so it would just be 1.8 then right 

Hmmm!

A sine function on its own has a maximum of 1 and a minimum of -1 with a midline of 0; so what value does f(x) have if you replace the sine function in it by 0?

There is a triangle ABC, and there are 3 equilateral triangles such that each side of ABC is the base of one equilateral triangle, and the equilateral triangles are facing outwards(is not overlapping ABC).

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I didn't know this. 

Melody did get it right :) But I'd like to point out an easier method.

Multiply by \(\frac{\sqrt[16]{5}-1}{\sqrt[16]{5}-1}\), and the expression simplifies to \(\sqrt[16]{5}-1\). Our answer is hence \((\sqrt[16]{5})^{48} = \boxed{125}\).

Wrong to the power of h**l

If the coefficient of x is 1, then the factorized form would be something like (x-a)(x-b) for roots a and b. In this case, the roots are -2 and 3, so it would be (x+2)(x-3)=$\boxed{x^2-x-6}$

15 * 43 = 645

39 * 14 = 546

645 - 546 = 99

Max = 99

The maximum number of beads is 14.

The domain is (-inf,-8) U (-8,2) U (2,inf).

Using your formula, the least possible value of k is -120.

The wording of the question is a bit messed up....please proofread ....

Nice, but I think you forgot 0, 0^2 = 0. 

=^._.^=

The domain of g(x) is [-16,2].

$f(g^{-1}(f^{-1}(f^{-1}(g(f(5)))))) = 1/4$.

Case 1: x - 1 = x + 2 =====> 0 = 3 (0 ≠ 3 so yh)

Case 2: x - 1 = -x - 2 =====>2x = -1 so x = -1/2

6*4 + 3*7 = 45 

A builder has some red and grey bricks. The red bricks weigh 6 kilograms each, and the grey bricks weigh 7 kilograms each. If the total weight of all 7 bricks is 46 kilograms, how many of each type of brick does Cooper have?
 

6*6 + 7 = 43

6*5 + 2*7 = 44

6*5 + 3*7 = 45

6*3 + 4*7 = 46 

What is the domain of the function $$u(x) = \frac{1}{\sqrt[3]{x}}~?$$ Express your answer in interval notation.

Hello Guest!

\(u(x) = \frac{1}{\sqrt[3]{x}}~?\)

\(D\in \mathbb R\ |\ x\neq 0\)

  !

By the quadratic formula, the projectile reaches 36 feet in 1.2 seconds.

it's +8/12  

of course that will reduce, but so would have 64/144 in the first place

oh lol

Guest, your English is a bit off. I put this into a Grammarly document, and I got your question now! 

The question, is Grammarly is correct, is: How many ways are there to choose french fries and potatoes if there are 5 french fries and 2 potatoes?

Well, you can choose 0, 1, 2, 3, 4, 5, or 6 french fries. You can also pick 0, 1, or 2 potatoes.

The number of choices are $7 \cdot 3 = \boxed{21}$