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BaLLatris [955]
2 years ago
11

A container has the shape of an open right circular cone, as shown. The height of the container is 10 cm and the diameter of the

opening is 10 cm. water in the container is evaporating so that its depth h is changing at a constant rate of -3/10cm/hr. (v=1/3pir^2h). A) find the volume when h=5cm and B) fine the rate of change of the volume of water when h=5
Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
5 0
A) Volume = (1/12)pi*h^3, with height = 5cm. 

<span>b) You should be able to differentiate V = (1/12)pi*h^3 with respect to h, and you were given dh/dt = -0.3 cm/hr. 
</span>
does that make sense?
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What is the question?
I'm assuming it is to find the length and width.
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8 0
2 years ago
f(x) = −16x2 + 24x + 16 Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points) Part B: Is the vertex
DerKrebs [107]

Answer:

see below

Step-by-step explanation:

f(x) = −16x^2 + 24x + 16

Set equal to zero to find the x intercepts

0  = −16x^2 + 24x + 16

Factor out -8

0 = -8(2x^2 -3x-2)

Factor

0 = -8(2x +1) (x-2)

Using the zero product property

2x+1 =0    x-2 =0

x = -1/2      x=2

The x intercepts are -1/2   ,2

Since the coefficient of x^2 is negative the graph will open down and the vertex will be a maximum

The x value of the maximum is 1/2 way between the zeros

(-1/2+2) /2 = 1.5/2 =.75

To find the y value substitute into the function

f(.75) = -8(2x +1) (x-2)

        =-8(2*.75+1) (.75-2)

        = -8(2.5)(-1.25)

         =25

The vertex is at (.75, 25)

We have the zeros, and  the vertex.  We know the graph is symmetrical about the vertex

3 0
2 years ago
In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline
VLD [36.1K]

Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>

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Cant figure this one out.

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