Please help I need help with this homework ASAP!!

Simplify the following.<br> c*5 x c

posledela

Answer:

c^6

Step-by-step explanation:

c^5*c^1=c^5+1=c^6

8 0

3 years ago

Please help me asap

german

The answer is 86.4 I believe.

5 0

3 years ago

Kindly solve ½=4/?, What's going to be the answer? Thank you

Luda [366]

Answer:

8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

To estimate the height of a house Katie stood a certain distance from the house and determined that the angle of elevation to th

Nata [24]

the height of the house is $408ft$ .

<u>Step-by-step explanation:</u>

Here we have , To estimate the height of a house Katie stood a certain distance from the house and determined that the angle of elevation to the top of the house was 32 degrees. Katie then moved 200 feet closer to the house along a level street and determined the angle of elevation was 42 degrees. We need to find What is the height of the house . Let's find out:

Let y is the unknown height of the house, and x is the unknown number of feet she is standing from the house.

Distance of house from point A( initial point ) = x ft

Distance of house from point B( when she traveled 200 ft towards street = x-200 ft

Now , According to question these scenarios are of right angle triangle as

At point A

⇒ $Tan32 =\frac{Perpendicular}{Base}= \frac{y}{x}$

⇒ $Tan32 = \frac{y}{x}$

⇒ $y=x(Tan32 )$ ..................(1)

Also , At point B

⇒ $Tan42 = \frac{y}{x-200}$

⇒ $y=(x-200)(Tan42)$ ..............(2)

Equating both equations:

⇒ $(x-200)(Tan42) = x(Tan32)$

⇒ $x(Tan42-Tan32)=Tan42(200)$

⇒ $x=\frac{Tan42(200)}{(Tan42-Tan32)}$

⇒ $x=653ft$

Putting $x=653ft$ in $y=x(Tan32 )$ we get:

⇒ $y=x(Tan32 )$

⇒ $y=653(Tan32 )$

⇒ $y=408ft$

Therefore , the height of the house is $408ft$ .

4 0

3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$