Im not sure but i think its A because you have no value to multiply 3 so you just multiply it by itself.It would be a negative because you cant divide 3 by 2.so it would be A.-x9y2.
i hope i was correct and helped if not you can go ahead and delete my answer:)
Answer:
12πx⁴, 15x⁷, 16x⁹
Step-by-step explanation:
Volume of a cylinder: πr²h
Volume of a rectangular prism: whl
Plugging in variables for the volume of a cylinder, we get: 3x²·(2x)²·π
3x²·(2x)² = 3·2·2·x·x·x·x
= 12·x⁴
=12x⁴
Now, we just multiply that by π.
12x⁴·π = 12x⁴π
A monomial is a 1-term polynomial, so 12x⁴π is a monomial.
Plugging in variables for the volume of a rectangular prism, we get: 5x³·3x²·x²
5x³·3x² = 5·3·x·x·x·x·x
= 15·x⁵
= 15x⁵
Now, we just multiply that by x².
15x⁵·x²
= 15·x·x·x·x·x·x·x
= 15·x⁷
=15x⁷
A monomial is a 1-term polynomial, so 15x⁷ is a monomial.
Same steps for the last shape, another rectangular prism:
2x²·2x³·4x⁴
2x²·2x³
= 2·2·x·x·x·x·x
= 4·x⁵
= 4x⁵
Now, we just multiply that by 4x⁴.
4x⁵·4x⁴
= 4·4·x·x·x·x·x·x·x·x·x·
= 16·x⁹
= 16x⁹
A monomial is a 1-term polynomial, so 16x⁹ is a monomial.
Hi there!
3/4 of sugar is needed for 6 servings.
1/8 x 6 = 6/8 simplify it and you get 3/4.
So 3/4 is your answer.
Hope this helps ;D
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is

. Set the derivative equal to 0 and factor to find the critical numbers.

, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.