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ss7ja [257]
3 years ago
9

Please help me will mark brainliest!!! Provide a hand drawn graph attachment pls

Mathematics
1 answer:
Travka [436]3 years ago
5 0

The graph is shown below. I used GeoGebra to create the graph. The graph is restricted on the domain 0 \le x \le 2 meaning that everything to the left of x = 0 is not drawn, and the same for everything to the right of x = 2.

A table of values is included as well. Each row in the table represents an ordered pair point (x,y) that is on the blue cosine graph.

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The Answer is:
5 and 3/4
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If f(x)=-4x+1 what is f(-x/2)
guapka [62]

Answer:

f(\frac{-x}{2})=2x+1

Step-by-step explanation:

given f(x)=-4x+1................(1)

x=\frac{-x}{2}  put in equ (1)

f(x)=-4x+1

f(\frac{-x}{2} )  =-4(\frac{-x}{2})+1

 f(\frac{-x}{2})=2x+1 answer

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Please help me ASAP!!
raketka [301]
-15u+30
Have a great day! :)
6 0
3 years ago
Read 2 more answers
Find a parametric representation for the part of the cylinder y2 + z2 = 49 that lies between the planes x = 0 and x = 1. x = u y
sweet-ann [11.9K]

Answer:

The equation for z for the parametric representation is  z = 7 \sin (v) and the interval for u is 0\le u\le 1.

Step-by-step explanation:

You have the full question but due lack of spacing it looks incomplete, thus the full question with spacing is:

Find a parametric representation for the part of the cylinder y^2+z^2 = 49, that lies between the places x = 0 and x = 1.

x=u\\ y= 7 \cos(v)\\z=? \\ 0\le v\le 2\pi \\ ?\le u\le ?

Thus the goal of the exercise is to complete the parameterization and find the equation for z and complete the interval for u

Interval for u

Since x goes from 0 to 1, and if x = u, we can write the interval as

0\le u\le 1

Equation for z.

Replacing the given equation for the parameterization y = 7 \cos(v) on the given equation for the cylinder give us

(7 \cos(v))^2 +z^2 = 49 \\ 49 \cos^2 (v)+z^2 = 49

Solving for z, by moving 49 \cos^2 (v) to the other side

z^2 = 49-49 \cos^2 (v)

Factoring

z^2 = 49(1- \cos^2 (v))

So then we can apply Pythagorean Theorem:

\sin^2(v)+\cos^2(v) =1

And solving for sine from the theorem.

\sin^2(v) = 1-\cos^2(v)

Thus replacing on the exercise we get

z^2 = 49\sin^2 (v)

So we can take the square root of both sides and we get

z = 7 \sin (v)

4 0
3 years ago
at noon on jan 7, the temperature in buffalo ny was 1 degree below of?. the temp in pittsbug, p at the same time was 10 degrees
Bumek [7]

Answer:

the temperature in Pittsburgh was 41 degrees Farenheit

4 0
3 years ago
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