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3 years ago

A cylinder has a height of 13 feet and a radius of 11 feet. What is its volume? Use a ~ 3.14

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

900

Step-by-step explanation:

V=πr2h

3.14 x 11= 34.54

34.54 x 2= 69.08

69.08x 13= 898.04

898.04 rounded to nearest hundreth is 900

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A plane flies over city A and reaches city B, 115 miles away, in 15 minutes. Find the speed of the plane in miles per hour assum

Vera_Pavlovna [14]

So you want to find how many miles are being flown in one hour.

There are 60 minutes in one hour so you want to find how many times 15 goes into 60 so you can then multiply that by 115.

60 / 15 = 4.

Now multiply 115 by 4,

115 x 4 = 460

Therefore the plane is flying at B. 460 mph.

Hope I helped!

3 0

3 years ago

For 10-18, Determine whether the triangles can be proved

navik [9.2K]

Answer:

180 - 92 - 41 = 47

There are not two similar congruent angle in the triangles.

Hope this helps!

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2 years ago

Plsss help i’ll give brainliest if you give a correct answer

Ymorist [56]

Answer:

300

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Look at the graph below measuring the speed of a truck and answers the following 2 question.

NemiM [27]

X would be equal to time and y be equal to distance

The y axis would be measuring distance and the x axis would be measuring time. Another name for a speed graph is the distance time graph. The y axis would be measuring the distance covered by the truck in meters and the x axis would be for the time it takes to reach that distance in second.

B) 2-6 would represent contact motion, meaning the vehicle is at rest. The time would be increasing from 2-6 to however the distance remains the same which is 20m

4 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

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

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

3 years ago