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LiRa [457]
3 years ago
6

An ash borer is an invasive pest whose larvae eat the pulp of ash trees as they mature. A park ranger has a tree that is infeste

d with ash borers. She estimates they have eaten approximately 40% of the tree's pulp. If the ash tree's trunk has a radius of 2 feet and a height of 15 feet, what was the total volume of the tree trunk before the ash borers started eating it?
Mathematics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

188.6 cubic feet

Step-by-step explanation:

Let r, h denotes radius and height of the tree's trunk.

Radius of the tree's trunk = 2 feets

Height of the tree's trunk = 15 feets

The tree's trunk is in the shape of a cylinder.

Volume of cylinder (tree's trunk) =\pi r^2h

Put r=2\,,\,h=15

Volume of the tree's trunk =\pi (2)^2(15)=60\pi cubic feet

Put \pi=\frac{22}{7}

So,

Volume of the tree's trunk =60(\frac{22}{7})=\frac{1320}{7}=188.6 cubic feet

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umka21 [38]
2 cakes - Theresa's and Joe's
Theresa's cake had 6 pieces after she cut it. (2 times the size of Joe's pieces)
Joe's cake had 12 pieces after he cut it. (1/2 the size of Theresa's pieces)

We know that 8/12ths of ONE cake were eaten and that Joe ate 2 of his pieces.
We want to know how many pieces Theresa ate of her cake. Keeping in mind that her pieces are equal to 2 of Joe's pieces we can solve this question. 

8/12 eaten total
if 2/12 by Joe
then 8-2 = 6, 6/12 by Theresa
(BUT: Theresa's pieces were twice the size of Joe's so we will divide by 2)
6/12 = 3/6

Answer: Theresa ate 3 pieces of her cake

7 0
3 years ago
What is the constant variation in the equation 3y=6x? Explain. PLEASE HELP IM DESPERATE
lbvjy [14]
Hi Desperate!

anyways the equation for constant variation is y=kx
3y=6x is it, but we don't want the 3.
so we divide on both sides to cancel out the 3.
y=2x
4 0
3 years ago
The following integral requires a preliminary step such as long division or a change of variables before using the method of par
shtirl [24]

Division yields

\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}

Now for partial fractions: you're looking for constants <em>a</em>, <em>b</em>, and <em>c</em> such that

\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}

\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a

which gives <em>a</em> + <em>b</em> = 2, <em>c</em> = 0, and 2<em>a</em> = -7, so that <em>a</em> = -7/2 and <em>b</em> = 11/2. Then

\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}

Now, in the integral we get

\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx

The first two terms are trivial to integrate. For the third, substitute <em>y</em> = <em>x</em> ² + 2 and d<em>y</em> = 2<em>x</em> d<em>x</em> to get

\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}

7 0
2 years ago
This figure is 14 of a circle.
Wittaler [7]

Answer:

The perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.

Step-by-step explanation:

Given : A quarter of a circle with radius labeled 2 in.

Quarter of circle means fourth part of a circle  (as shown in image by shaded part).

We have to find the perimeter of this shaded quarter of circle.

Perimeter is the sum of the dimension of the given figure.

Perimeter of circle is 2\pi r

Since, we are given Quarter circle so its perimeter \frac{2}{4} \pi r+r+r=\frac{1}{2} \pi r+r+r

Perimeter of given figure = \frac{1}{2} \pi r+r+r

Perimeter of given figure = \frac{1}{2}\cdot 3.14 \cdot 2+2+2=3.14+4=7.14

Thus, the perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.

6 0
3 years ago
Read 2 more answers
Point A is located at (1,5), and point M is located at (-1, 6). If point M is the midpoint of AB, find the
Step2247 [10]

Answer:

-3, 7

Step-by-step explanation:

because you take how much from the first coordinate to the second and add it too the second

7 0
3 years ago
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