All categories

4 years ago

a cylinder has radius 1.6 meters. it’s volume is 95 cubic meters. find it’s height to the nearest tenth of a meter

Mathematics

1 answer:

erastova [34]4 years ago

6 0

H = 11.81 m

v= π r2 h

h = v/ π r2 = 95/ π 1.6(2)* = 11.81228m or just 11.81

*(2) is squared

hope this helped haha

You might be interested in

2. Two of the radii for the inscribed circle below have the expression 2x -8 and 4x

expeople1 [14]

Answer:

Step-by-step explanation:

6 0

3 years ago

A bus travels two different routes: the Green Route and the Blue Route.

Alex17521 [72]

Answer:4.5

Step-by-step explanation: Check the file for my explanation

8 0

2 years ago

I will give brainliest!

svetlana [45]

A copy of a line segment will have the same measure or length as the segment.

The true statement is (4) They are congruent.

From the question, we understand that a copy of the line segment is created.

This means that the new segment has the exact property as the original line segment.

Hence, both lines are congruent, and the true statement is (d)

Read more about line segments at:

brainly.com/question/3573606

6 0

3 years ago

Help please =*7=76=)(=(*+*(=

Triss [41]

Answer:

Likely

Step-by-step explanation:

8 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

3 years ago