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3 years ago

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Mathematics

1 answer:

Virty [35]3 years ago

8 0

Answer:

(1) 23.55mm³

(2) 340.17in³

Step-by-step explanation:

Volume of a cone is: $V=\frac{1}{3} \pi r^2h$

'r' - radius

'h' - height

The height of Cone 1 is 10 mm. The Diameter of Cone 1 is 3mm.

Radius is half the Diameter.

$3/2=1.5$

The radius is 1.5 mm.

$\text {Using 3.14 as pi: }\\V=(1/3)3.14*1.5^2*10\\V=(1/3)3.14*2.25*10\\V=(1/3)70.65\\\boxed{V=23.55}$

The volume of Cone 1 should be about 23.55mm³.

The height of Cone 2 is 13 in. The Radius of Cone 2 is 5 in.

Using 3.14 as pi:

$V=(1/3)3.14*5^2*13\\V=(1/3)3.14*25*13\\V=(1/3)1020.5\\\boxed {V=340.166666667}$

340.166666667 ≈ 340.17

The volume of Cone 2 should be about 340.17in³.

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First you find the median, by taking 893+954 and dividing the sum of those two numbers by 2. The median is 923.5. Then you take and find the median of the numbers before the median which is 513. That is your lower quartile. Then you find the median of the numbers below the median, which is 1038. That is your upper quartile. The interquartile range is your upper quartile minus your lower quartile.

Therefore, your answer would be:
Median- 923.5
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3 years ago

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3 years ago

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Svetach [21]

Answer:

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Step-by-step explanation:

Let

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y----> the radius of smaller circle

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3 years ago

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3 years ago

Jennifer hit a golf ball from the ground and it followed the projectile ℎ(t)= −15t^2+100t, where t is the time in seconds, and ℎ

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

In order to find the max height the ball reached, we have to complete the square on that quadratic. That will also, conveniently so, give us the number of seconds it will take the ball to reach that max height, that answer to part b. Let's begin to complete the square. Normally, you would move the constant over to the other side of the equals sign, but there is no constant here. The next step is to get the leading coefficient to be a 1, and ours right now is a -15. So we have to factor it out. Here's where we start the process of completing the square.

$-15(t^2-\frac{20}{3}t)=0$ Next step is to take half the linear term, square it, and add it to both sides. Our linear term is 20/3. Half of 20/3 is 20/6, and 20/6 squared is 400/36.

$-15(t^2-\frac{20}{3}t+\frac{400}{36})=0+???$ Because this is an equation, what we add to the left side also has to be added to the right. BUT we didn't just add in 400/36, because we have that -15 out front as a multiplier that refuses to be ignored. What we actually added in was -15(400/36):

$-15(t^2-\frac{20}{3}t+\frac{400}{36})=0-\frac{500}{3}$

The reason we do this is to create a perfect square binomial on the left which will serve as the number of seconds, h, in the vertex (h, k), where h is the number of seconds it takes the ball to reach its max height, k. Simplifying both sides then gives us:

$-15(t-\frac{20}{6})^2=-\frac{500}{3}$ Finally, we will move the right side over by the left and set the quadratic back equal to h(t):

$h(t)=-15(t^2-\frac{20}{3})^2+\frac{500}{3}$ and from that you can determine that the vertex is $(\frac{20}{3},\frac{500}{3})$ .

The answer to a. is vound in the second number of our vertex: k, the max height. The max of the golf ball was 500/3 feet or 166 2/3 feet.

Part b is found in the first number of the vertex: h, the number of seconds it took the golf ball to reach that max height. The time it took was 3 1/3 seconds.

Part c. is to state the domain (the time) and the range (the height) of the ball.

Domain is

D: {x | 0 ≤ x ≤ 3 1/3} and

Range is

R: {y | 0 ≤ y ≤ 166 2/3}

8 0

3 years ago

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