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malfutka [58]
3 years ago
8

Which equation is the inverse of y=x^2-36

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

y=\sqrt{x+36}

Step-by-step explanation:

Since, for finding the inverse of a function f(x) we follow the following steps,

Step 1 : Replace f(x) by y,

Step 2 : Interchange x and y,

Step 3 : isolate y in the left side,

Step 4 : Replace y by f^{-1}(x)

Given equation is,

y=x^2-36

Here, we do not need to apply step 1 and step 4,

Now, Interchange x and y,

x=y^2-36

-y^2=-x-36

y^2=x+36

Taking square root both sides,

y=\sqrt{x+36}

Which is the required equation of inverse of the given equation.

Ad libitum [116K]3 years ago
5 0
For these kinds of questions you should first solve for x:
y +36 = x^2 ⇒ √y+36 = x 
now the inverse of this equation is:
√x + 36 = y :)))
i hope this is helpful
have a nice day 
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True or false?
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Answer:

False: A function is a rule that assigns each vale of the independent variable to exactly one vale of the dependent variable.

Step-by-step explanation:

4 0
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Mr. Fuller wants to put fencing around his rectangular-shaped yard. the width of the yard is 55 feet and the length is 75 feet.
AlekseyPX
260 feet. 55+55=110, 75+75=150, 150+110=260
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2 years ago
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Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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cylindrical vase holds 8,038.4 cubic centimeters of water. The height of the vase is 40 centimeters. What is the length of the r
iragen [17]

The radius of the cylindrical vase is 8 centimeters, if the cylindrical vase holds 8,038.4 cubic centimeters of water and the height of the vase is 40 centimeters.

Step-by-step explanation:

The given is,

                 Volume of cylindrical vase = 8038.4 cubic centimeters

                 Height of the cylindrical vase = 40 centimeters

Step:1

                Formula for volume of cylindrical vase,

                                     Volume, V=\pi r^{2} h......................(1)

               Where, r - Radius of cylindrical vase

                           h - Height of cylindrical vase  

                From the given,

                                      V = 8038.4 cubic centimeters

                                      h = 40  centimeters  

                Equation (1) becomes,

                             8038.4 = \pi r^{2}  40

                                          =(3.14)(40)r^{2}                               (∵ \pi = 3.14 )

                               8038.4 = 125.6 r^{2}

                                      r^{2} = \frac{8038.4}{125.6}            

                                       r =\sqrt{64}

                                       r = 8 centimeters

Result:

           The radius of the cylindrical vase is 8 centimeters, if the cylindrical vase holds 8,038.4 cubic centimeters of water and the height of the vase is 40 centimeters.

7 0
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