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3 years ago

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A bird feeder is in the shape of a cylinder. It has a volume of about 100 cubic inches. It has a radius of 2 inches. What is the

approximate height of the bird feeder? Use 3.14 for pi. Math item stem image CLEAR CHECK 4 inches 8 inches 12 inches 79 inches

2 answers:

Lilit [14]3 years ago

8 0

8 inches because 3.14*2^2*8=100.48 which is rounded to 100

son4ous [18]3 years ago

3 0

Answer:

The approximate height of the bird feeder is $8\ in$

Step-by-step explanation:

we know that

The volume of a cylinder is equal to

$V=\pi r^{2}h$

In this problem we have

$V=100\ in^{3}$

$r=2\ in$

substitute and solve for h

$100=(3.14)(2^{2})h$

$100=12.56h$

$h=100/12.56=7.96=8\ in$

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Find the value of x for which the graph of y = 3x^2 - 8x + 7 achieves its minimum y-value.

gulaghasi [49]

Answer:

<h3>The y value achieves its minimum at x = 4/3</h3>

Step-by-step explanation:

Given the graph of y to be 3x² - 8x + 7, to get the value of x for which the graph function achieves its minimum y value, we need to find its turning point first.

At the turning point, dy/dx = 0

Given y = 3x² - 8x + 7

$\frac{dy}{dx} = 6x-8\\ at\ turning\ point\ 6x-8 = 0$

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What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

8 0

2 years ago

The length of a pipe that weighs 45 pounds is cut into 2 pieces. One piece is 3 feet long and weighs 18 pounds. What is the leng

muminat

If the length of a uniform pipe is 45 pounds, then the 18 pound piece is 18/45 the length of the pipe. If 18/45 of the length of the pipe is equal to 3 feet, then we can set up the equation:
(18/45)x = 3ft
Where x is the total length of the pipe. Solving for x:
<span>(18/45)x = 3ft
</span>18x = 135
x = 7.5 feet

We now know that the total length of the pipe before being cut is 7.5 feet. Since the first piece was 3 feet, we know that the second piece must be 4.5 feet (7.5 - 3).

Answer: 4.5 feet

6 0

3 years ago