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STatiana [176]
3 years ago
12

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 4 cubic feet

per minute. If the pool has radius 4 feet and height 10 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 8 feet?
Mathematics
1 answer:
lesantik [10]3 years ago
5 0
Final answer is 49 \pi /5


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Parallel Lines A line is parallel to y = 2x - 8 and intersects the point (-4,-1). What is the equation of this parallel line?
Assoli18 [71]

Answer:

y= 2x+7

Step-by-step explanation:

Since the line is parallel to y = 2x - 8 it has slope m= 2

The equation of the line passing through (-4,-1) is given by,

\frac{y - ( - 1)}{x - ( - 4)}  = m

\frac{y  + 1}{x + 4}  = 2

y  + 1 = 2(x + 4)

y + 1 = 2x + 8

y = 2x + 8 - 1

y = 2x + 7

4 0
3 years ago
Darin has 5 coins that the total 62%. What are the coins?
Snowcat [4.5K]
Hi there!

There are several possible options.

A main one would be : 10 DIMES and 2 PENNIES

Other ones could be :

2 quarters
1 dime
2 pennies

62 pennies

12 nickels
2 pennies

and more !

Hope this helps !
5 0
3 years ago
Which answer describes the graph of the system of equations?
Vladimir [108]

Answer: The point of intersection is (0,4).

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Let f(x)=(18)3^x. find the equation of the line between the points (-2,f(-2)) and (1,f(1))
horsena [70]

\text{Given that,} \\\\f(x) = 18 \cdot 3^x \\\\f(-2) = 18 \cdot 3^{-2} = 2\\\\f(1) = 18 \cdot 3^1 = 54\\ \\\text{So,}~ (x_1,y_1) = (x_1, f(-2)) = (-2,2) ~\text{and}~ (x_2,y_2) = (x_2, f(1)) = (1,54)\\ \\\text{Slope,}~ m = \dfrac{y_2 -y_1}{x_2 -x_1} = \dfrac{54-2}{1+2} = \dfrac{52}3 \\\\

\text{Equation of line,}\\\\~~~~~~~y - y_1 = m (x-x_1)\\\\\\\implies y-2 = \dfrac{52}3(x+2)~~~~~~~~~~~;[\text{Point-Slope form.]} \\\\\\\implies y-2 = \dfrac{52}3x + \dfrac{104}3\\\\\\\implies y = \dfrac{52}{3}x + \dfrac{104} 3 +2\\\\\\\implies y = \dfrac{52}3 x +\dfrac{110}3 ~~~~~~~~~~~~~~~;[\text{Slope -Intercept form}]

6 0
2 years ago
What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?
xxMikexx [17]

We have been given an equation of hyperbola \frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1. We are asked to find the center of hyperbola.  

We know that standard equation of a vertical hyperbola is in form \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1, where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1

Now we can see that value of k is -3. Therefore, the vertex of given hyperbola will be at point (6,-3) and option D is the correct choice.

6 0
3 years ago
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