Answer:
h = 16x
Step-by-step explanation:
The volume of a sphere and a cone is same such that,

We have, r = x, R = (1/2)x, h= ? (height of the cone)
So,

So, the height of the cone is equal to 16x.
Answer:
Minimum at (-4, -10)
Step-by-step explanation:
x² + 8x + 6
The coefficient of x² is positive, so the parabola opens upward, and the vertex is a minimum.
Subtract the constant from each side
x² + 8x = -6
Square half the coefficient of x
(8/2)² = 4² = 16
Add it to each side of the equation
x² + 8x + 16 = 10
Write the left-hand side as the square of a binomial
(x + 4)² = 10
Subtract 10 from each side of the equation
(x+ 4)² -10 = 0
This is the vertex form of the parabola:
(x - h)² + k = 0,
where (h, k) is the vertex.
h = -4 and k = -10, so the vertex is at (-4, -10).
The Figure below shows your parabola with a minimum at (-4, -10).
The equation of the line is C. 1X/2 -2
How you figure this out is you first say when x is zero y is what?
y= 1/2 (0) - 2
You would then see that y=-2 (0,-2)
Plot that point
Next you need to find the slope of the line.
Remember slope is rise over run.
When you 'rise' up one (from -2) and 'run' 2
you are now at the point (2,-1) which is on the line.
You can continue doing rise/run along the line to check your answer.
I hope this makes sense! Let me know if you have any questions :)
Answer:
5 + 5 = 10
5 + 6 = 11
5 + 7 = 12
5 + 8 = 13
5 + 9 = 14
5 + 10 = 15
5 + 11 = 16
Step-by-step explanation:
Answer:
%82.5
Step-by-step explanation:
- The final exam of a particular class makes up 40% of the final grade
- Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.
From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:
,
where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,
.
So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:

That is, he need al least %82.5 to obtain a passing grade.