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3 years ago

5

Galila a sprinkler that sprays water in a circle. The circle has a 20-foot radius. Which of the following equations can be used

to determine the circumference of the circle created by sprinkler water?

1 answer:

Mariana [72]3 years ago

7 0

C = pi x d or c = 2 x pi x r

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Find f(2) given f(x) = -3x^3 + x^2 – 3

aleksley [76]

Answer:

D

Step-by-step explanation:

f(x) = -3x^3 + x^2 – 3 f(2) means that wherever you see a x, put in a 2.

f(2)= -3(2)^3 + (2)^2 - 3

f(2) = -3*8 + 4 - 3

f(2) = - 24 + 1

f(2) = - 23

6 0

2 years ago

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Given the 3rd term =3.2 and d=-0.4, find the fourth term.

Marat540 [252]

That is the solution

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3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago

One thumbtack weighs 0.25 ounce. How many thumbtacks are in a box that has a mass of 3500 grams? (Hint: 1oz ~28g

Marina86 [1]

Alright so .25 oz is 1/4 of 1oz. Therefore you take 28 and divide it by 4 and get 7.

Then take 3500 and divide it by 7 which makes 500 thumb tacks.

I think idk though.

7 0

3 years ago

(3) Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and t

Gekata [30.6K]

Answer:

The rate at which the distance between the two people is changing 15 seconds later is 7.28 ft/sec.

Step-by-step explanation:

Here, we note the following

Let the person walking way fom the elevator be X

Let the other person going up in the elevator be Y

Therefore after 15 seconds, their positions will be;

For X, 2 ft/sec × 15 s = 30 ft away from the elevator

For Y, 7 ft/sec × 15 s = 105 ft up in the elevator

At that instant, the distance between them is given as

d² = x² + y²

= 30² +105² = 11925 ft²

d = √11925 ft² = 109.202 ft

The rate of change of the distance between the two peopl, X and Y is given as

$\frac{dd^2}{dt} = \frac{dx^2}{dt} + \frac{dy^2}{dt}$

Therefore, $2d\frac{dd}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ or $d\frac{dd}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$

Since $\frac{dx}{dt}$ is given as 2 ft/sec and $\frac{dy}{dt}$ = 7 ft/sec

Then $d\frac{dd}{dt} = 30 \times 2 + 105 \times 7 = 795\\$

$\frac{dd}{dt} = \frac{795}{109.202 } =7.28 ft/sec$

8 0

3 years ago

Read 2 more answers