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Harlamova29_29 [7]
3 years ago
12

You are downloading a song. The percent y(in decimal form) of megabytes remaining to download after x seconds is y=-0.1x+1.

Mathematics
2 answers:
Xelga [282]3 years ago
5 0
The Y intercept is always when X equals 0. The X intercept is when Y equals 0.
An easy way to find x and y intercepts is to substitute these variables with 0.
To find the Y intercept, substitute 0 for X.
Y=-.1(0)+1=1 So you know that when Y equals 1 and X equals 0, you have you Y intercept.
Because we write coordinates as (X,Y), you can just plug in the numerals and have your coordinate for the Y intercept. So, x=0 y=1 (0,1)
To find the X intercept, substitute 0 for Y.
     0=-.1x+1 -Add .1x to both sides
+.1x   +.1x
   .1x=1       -Divide both side to get x by itself.
.1x/.1=1/.1
x=10   So, we know that when y=0, x=10. This is our x intercept.
Coordinate form is (X,Y) so our x intercept is (10,0)
igomit [66]3 years ago
3 0


I would make the equation have both of your variables on the same side:

0.1x+y=1

Then I would use the cover-up method and say that

x-intercept= 10

y-intercept= 1


*Cover-Up Method: If you were finding the X-intercept, you would just ignore the y variable and divide your whole number by x.

In this case the whole number is 1. 1/0.1=10

Vice Versa with finding the Y-intercept

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Prove it. this question is from trigonometry ​
konstantin123 [22]

Answer:

See below for step-by-step proof.

Step-by-step explanation:

\textsf{Given expression}:

\cos^6 \theta+\sin^6 \theta

\textsf{Apply exponent rule} \quad a^{bc}=(a^b)^c:

\implies (\cos^2 \theta)^3+(\sin^2 \theta)^3

\textsf{Use the identity}\quad a^3+b^3=(a+b)^3-3ab(a+b):

\implies (\cos^2 \theta+\sin^2 \theta)^3-3 \cos^2 \theta\sin^2 \theta (\cos^2 \theta+\sin^2 \theta)

\textsf{Use the identity}\quad \sin^2 \theta + \cos^2 \theta=1:

\implies (1)^3-3 \cos^2 \theta\sin^2 \theta (1)

\implies 1-3 \cos^2 \theta\sin^2 \theta

\implies 1-3 (\cos \theta\sin \theta)^2

\textsf{Use the identity}\quad \sin 2 \theta=2 \sin \theta \cos \theta \implies \dfrac{1}{2}\sin 2 \theta= \sin \theta \cos \theta:

\implies 1-3 \left( \dfrac{1}{2}\sin 2 \theta\right)^2

\implies 1-\dfrac{3}{4} \sin^2 2 \theta

\textsf{Use the identity}\quad \sin^2 2\theta + \cos^2 2\theta=1 \implies \sin^2 2\theta=1-\cos^2 2\theta

\implies 1-\dfrac{3}{4} \left(1-\cos^2 2 \theta\right)

\implies 1-\dfrac{3}{4} +\dfrac{3}{4}\cos^2 2 \theta

\implies \dfrac{1}{4} +\dfrac{3}{4}\cos^2 2 \theta

\textsf{Factor out }\dfrac{1}{4}:

\implies \dfrac{1}{4}\left(1+3\cos^2 2 \theta\right)

\textsf{Use the identity}\quad \cos (4 \theta)=2 \cos^2 2\theta - 1 \implies \cos^2 2 \theta=\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}:

\implies \dfrac{1}{4}\left(1+3\left(\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}\right)\right)

\implies \dfrac{1}{4}\left(1+\dfrac{3}{2}\cos 4 \theta+\dfrac{3}{2}\right)

\implies \dfrac{1}{4}\left(\dfrac{5}{2}+\dfrac{3}{2}\cos 4 \theta\right)

\implies \dfrac{1}{4} \cdot \dfrac{1}{2}\left(5+3\cos 4 \theta\right)

\implies \dfrac{1}{8}\left(5+3\cos 4 \theta\right)

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Answer:

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Sphinxa [80]

Answer:

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