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2 years ago

Pythagorean Theorem

Mathematics

2 answers:

tatiyna2 years ago

8 0

The answer is 10.6 :)

german2 years ago

8 0

Answer: a = 10.6

Pythag is a squared + b squared = c squared
We have c and b so we can sub those in to get a squared + 12 squared = 16 squared
Which equals a squared + 144 = 256
Then we - 144 from both sides to get a squared = 112
Then we square root 112 to get a = 10.6
Hope that helps

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Help plz:))) I’ll mark u brainliest <br> ASAP!!!

8090 [49]

Answer:

x=16

Step-by-step explanation:

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8 0

2 years ago

Which expressions are equivelant ti the one below? 16^x/4^x

Jlenok [28]

Answer:

$4^{x}$

Step-by-step explanation:

$\frac{16^x}{4^x}$ Divide the fraction. Since both have an exponent of x, the answer has an exponent of x.

$4^{x}$

If this answer is correct, please make me Brainliest!

4 0

3 years ago

A short-wave radio antenna is supported by two guy wires, 150 ft and 180 ft long. Each wire is attached to the top of the antenn

Dafna1 [17]

The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)

<h3>How to determine the distance between the two anchor points of the guy wire</h3>

The problem will be solved using SOH CAH TOA

let the distance between the 150 ft long guy wire and the radio antenna be x

let the distance between the 180 ft long guy wire and the radio antenna be y

cos 65° = x / 150

x = cos 65° * 150

x = 63.39 ft

The height of the antenna

sin 65° = height of antenna / 150

height of antenna = sin 65 * 150

height of antenna = 135.95

using Pythagoras theorem

(length of guy wire)² = (height of the antenna)² + (anchor distance)²

(anchor distance)² = 180² - 135.95²

anchor distance = √(180² - 135.95²)

anchor distance = 117.97

The anchor points distance apart

= 63.39 + 117.97

= 181 (to the nearest foot)

Learn more on Pythagoras theorem here:

brainly.com/question/29241066

#SPJ1

4 0

1 year ago

What are the solutions to the following system of equations? Select the correct answer below.

Fed [463]

I would say probably b or d

7 0

3 years ago

A chemical flows into a storage tank at a rate of (180+3t) liters per minute, where t is the time in minutes and 0<=t<=60

Yuliya22 [10]

Answer:

The amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

Step-by-step explanation:

Consider the provided information.

A chemical flows into a storage tank at a rate of (180+3t) liters per minute,

Let $c(t)$ is the amount of chemical in the take at <em>t </em>time.

Now find the rate of change of chemical flow during the first 20 minutes.

$\int\limits^{20}_{0} {c'(t)} \, dt =\int\limits^{20}_0 {(180+3t)} \, dt$

$\int\limits^{20}_{0} {c'(t)} \, dt =\left[180t+\dfrac{3}{2}t^2\right]^{20}_0$

$\int\limits^{20}_{0} {c'(t)} \, dt =3600+600$

$\int\limits^{20}_{0} {c'(t)} \, dt =4200$

So, the amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.

5 0

3 years ago