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3 years ago

The lateral area of a cone is 555 pi cm^2. The radius is 37 cm. What is the slant height to the nearest tenth of a centimeter?

Mathematics

2 answers:

andre [41]3 years ago

7 0

Answer:

Slant height = 15

Step-by-step explanation:

Well the lateral area of a cone is equal to πrl, where r = radius and l = slant height. So if the radius is 37, you will have the equation:

37π * l = 555π

Start by crossing out π

37 * l = 555

l = 15

This might be wrong since it is so clean, and the question asks to simplify.

sammy [17]3 years ago

5 0

Answer:

i would help you if i could but im sorry i dont know

Step-by-step explanation:

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Solve the simultaneous equations: 5x + 3y = 9 7x - 2y = 25 x=__ y=__

kondor19780726 [428]

Answer:

x= 3, y= -2

Step-by-step explanation:

5x+3y= 9

x= 25÷7 + 2÷7 y

5(25÷7 +2÷7 y)

5( 25÷7 +2÷7 y ) + 3y= 9

y= -2

x= 25÷7 + 2÷7 x (-2)

x= 3

5 0

2 years ago

A water tank that holds 30 gallons of water when full is leaking 3/4 gallon of water every hour. If no one notices and stops the

viva [34]

It will take 40 hours for the water tank to become empty.

Explanation:
Divide 30 by 0.75 (equivalent to 3/4).

7 0

2 years ago

At their family's pizzeria, Dave makes 6 pizzas in the first hour it is open and 3 pizzas each hour after that. Lauren makes 11

Nady [450]

Answer:

47 pizzas

Step-by-step explanation:

Dave makes 6 pizzas in the first hour it is open and 3 pizzas each hour after that.

If the number of hours after the first hour is represented as x, then the number of pizzas he makes after x hours is:

6 + 3x

Lauren makes 11 pizzas in the first hour and 3 pizzas each hour after that. Then:

11 + 3x

The pizzeria is open for 6 hours in all, that is, 5 hours after the first hour.

This means that:

x = 5

The total amount of pizzas that Dave and Lauren make is:

6 + 3x + 11 + 3x

=> 17 + 6x

Therefore, the total amount of pizza they make 5 hours after the first hour is:

17 + 6(5) = 17 + 30 = 47 pizzas

3 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

2 years ago

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s

stira [4]

Hello,

(x-4)(x+8)(x-2)(x-5i)(x+5i)
=(x^2+4x-32)(x-2)(x^2+25)
=(x^3+2x^2-40x+64)(x^2+25)
=x^5 +2x^4 -15x^3 +114 x^2-1000x+1600

4 0

3 years ago