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3 years ago

Choose the surface area of the pyramid

Mathematics

1 answer:

pav-90 [236]3 years ago

8 0

Answer:

64 cm^2

Step-by-step explanation:

2*6 = 12/2 = 6

6 * 8 = 48

48+base= 48+16 aka. 4*4

64 cm squared

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When 200 gallons of oil were removed from a tank, the volume of oil left in the tank was 3/7 of the tank's capacity. what was th

ki77a [65]

Answer:

Step-by-step explanation:

"When 200 gallons of oil were removed from a tank" algebraically looks like this:

V - 200.

"...the volume of oil left in the tank was 3/7 of the tank's capacity" algebraically looks like this:

3/7(V)

Therefore, the equation is

V - 200 = 3/7(V)

Begin by multiplying both sides by 7:

7(V - 200) = 3V and

7V - 1400 = 3V so

-1400 = -4V so

V = 350 gallons

That's if the volume of the oil in the tank was 3/7 of the tank's capacity.

For the other part of the problem, we set up the equation almost the same, except the 3/7 is a 1/2:

V - 200 = 1/2(V)

Multiplying both sides by 2 gives you

2(V - 200) = V and

2V - 400 = V so

-400 = -V so

V = 400

7 0

3 years ago

|3x+6|+3=6<br> What is the answer?

GrogVix [38]

Answer:

x=-3 or x=-1

Step-by-step explanation:

Subtract 3 from both sides of your equation, it should look like this |3x+6|+3-3=6-3.

Simplify to get |3x+6|=3

Apply the absolute rule to get 3x+6=-3 or 3x+6=3.

3x+6=-3 : x=-3

3x+6=3 : x=-1

I'm glad I was able to help have a great day!

8 0

3 years ago

Which graph equals ƒ(x) = -x + 6?

KIM [24]

The last graph is the answer

6 0

3 years ago

Read 2 more answers

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

3 years ago

A wire is to be attached to support a telephone pole. Because of surrounding buildings, sidewalks, and roadways, the wire must

ad-work [718]

This situation is a right triangle with the base measuring 18 and the hypotenuse 22. We are looking for x, the height up the pole that the wire is going to be attached to. We will use Pythagorean's Theorem to find the missing length. $22^2-18^2=x^2$ and $484-324=x^2$ . x^2 = 160 so x = $4 \sqrt{10}$ or 12.65 feet

3 0

3 years ago