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katovenus [111]
3 years ago
10

The formula for the volume of a cone is V=1/3(3.14)r^2(h) where V represents the volume, r represents the radius of the base, an

d h represents the height. What is the height of a cone with a volume of 66 cubic centimeters and a base with a radius of 3 centimeters?
F. 21 cm
G. 69.14 cm
H. 7 cm
J. 0.78 cm
Mathematics
1 answer:
jasenka [17]3 years ago
7 0

The original formula for the volume of a cone is

V=\dfrac{1}{3}\pi r^2h

O.K. You give the approximate value of pi in the formula.

V=\dfrac{1}{3}\cdot3.14r^2h

We have:

V=66\ cm^3,\ r=3\ cm

Substitute:

\dfrac{1}{3}\cdot3.14\cdot3^2h=66\\\\3.14\cdot3h=66\\\\9.42h=66\ \ \ \ |:9.42\\\\h\approx7\ cm

Answer: 7 cm.

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There are 20 questions in a quiz. Each correct answer scores 7 points, each wrong answer scores -4 points and each question left
Dovator [93]
<h3>Answer:    1  question left blank</h3>

He answered 16 questions correctly, and got 3 wrong answers.

==========================================================

Explanation:

Let

  • x = number of questions that are correct
  • y = number of wrong answers
  • z = number of questions left blank

x,y,z are nonnegative whole numbers.

-------------------

Since there are 20 questions total, this means the first equation to set up is:

x+y+z = 20

Solving for y leads to

x+y+z = 20

y+z = 20-x

y = 20-x-z

We'll use this later.

-------------------

Another equation to set up is 7x-4y = 100 because Eric earns 7 points per correct answer and loses 4 points for each incorrect answer, and all that leads to 100 points total which was his quiz score. We'll ignore the questions he left blank since they add 0 points.

Let's plug the equation in which we isolated y

7x-4y = 100  

7x-4(20-x-z) = 100

7x-80+4x+4z = 100

7x+4x+4z = 100+80

11x+4z = 180

-------------------

Now we can guess and check to see which pair of x and z values will make that last equation above true. I suggest starting with the smallest possible value of x and using that x value to solve for z.

If x = 0, then,

11x+4z = 180

11(0)+4z = 180

4z = 180

z = 180/4

z = 45

So if Eric got 0 correct answers, then he left 45 questions blank. But that isn't possible because there are only 20 questions total. So we'll ignore the case that x = 0.

If we increase x by 4, and decrease z by 11, then we get another ordered pair solution to this equation

So another solution is (x,z) = (4,34)

Note that

11x+4y = 180

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But like before, z = 34 isn't possible since 20 is the max.

Increase x by 4 again, and drop z by 11 to get (x,z) = (8,23). Again we run into the same issue as before.

Increase x by 4 again, and decrease z by 11 to get (x,z) = (12, 12). Now we have both x and z smaller than 20, but note how x+z = 12+12 = 24 which exceeds the total number of questions. So we rule this case out as well.

Do another round of "increase x by 4, decrease z by 11" to get to (x,z) = (16, 1). This is the only case left because anything beyond this, z will be negative.

Luckily, this final case does work. If Eric answers x = 16 questions correctly, then he left z = 1 of them blank. That must mean y = 20-x-z = 20-16-1 = 3 questions were incorrect.

We can see that:

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meaning that (x,y) = (16,3) is a solution to 7x-4y = 100.

-------------------

To summarize, we found that the only possible solution is (x,y,z) = (16, 3, 1)

Meaning x = 16 questions were correct, y = 3 were wrong, and z = 1 question was left blank.

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