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3 years ago

12

find the volume of a cone with a base diameter of 12in and a height of 9in . write the exact volume in terms of π , and be sure

to include the correct unit in your answer.

1 answer:

mariarad [96]3 years ago

4 0

The volume (V) of the cone is 1/3 of the product of the area (A) of the base and height (h). The area of the circular base is, A = πd²/4. So,
V = πd²h/ 12
Substituting the given values,
V = π(12 in)²(9 in) / 12 = 108π in³
Thus, the volume of the cone is 108π in³.

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3 years ago

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4 years ago

If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then is the triangle isosceles?

Ne4ueva [31]

Answer:

True

Step-by-step explanation:

The perpendicular bisector of the opposite side to the vertex bisects the angle at the vertex into two equal parts and also bisects the triangle into two equal parts.

Let A be the angle at the vertex, then assume that the angle is an isosceles triangle with base angles B.

We need to show that A = 180 - 2B for an isoceles triangle

The perpendicular bisector bisects A into two so the new angle in the vertex one half of the bisected triangle is A/2.

Since this half triangle is a right-angled triangle, the third angle in it is 90.

So, A/2 + B + 90 = 180 (Sum of angles in a triangle)

subtracting 90 from both sides, we have

A/2 + B + 90 - 90 = 180 - 90

A/2 + B = 90

subtracting B from both sides, we have

A/2 + B = 90

A/2 = 90 - B

multiplying through by 2, we have

A = 2(90 - B)

A = 180 - 2B

Since A = 180 - 2B, then our triangle is an isosceles triangle.

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3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago