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3 years ago

Given that f(x) = 2x − 5, find the value of x that makes f(x) = 15.

Mathematics

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

Step-by-step explanation:

so you replace all the x's with 15 and it becomes

2×15-5

2×15=30

30-5=25

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3 years ago

Choose the function whose graph is given by:

katen-ka-za [31]

Answer:

D. $y=\tan (x-\pi)+2$

Step-by-step explanation:

We are given the graph of the transformed tangent function.

As, we can see that the graph is shifted 2 units upwards.

That is, the parent function is translated 2 units up.

So, the new function will be of the form, $y=\tan (x)+2$ .

Thus, options A and C are not correct.

Moreover, 'If a function f(x) has period P, then the function cf(bx) will have period $\frac{P}{|b|}$ .

Also, form the figure, we see that the period of the graphed function is $\pi$ .

Since, $y=\tan (x)$ have period $\pi$ .

So, function in option B will have period $\frac{\pi}{2}$ , which is not possible.

<em>Further, function in option D have period $\pi$ .</em>

Thus, the equation of the graphed function is, $y=\tan (x-\pi)+2$ .

7 0

3 years ago

Read 2 more answers

Will give brainliest plz help

Dominik [7]

Answer:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Step-by-step explanation:

Given:

Center of the ellipse is, $(h,k)=(-4,-3)$

Minor axis length is, $2b=6$

A vertex of the ellipse is at (1, -3)

Now, distance between the center and the vertex is half of the length of the major axis.

Using distance formula for (-4, -3) and (1, -3), we get:

$a=\sqrt{(1+4)^2+(-3+3)^2}=5$

Therefore, the value of half of major axis is, $a=5$ . Also,

$2b=6\\b=\frac{6}{2}=3$

Now, equation of an ellipse with center $(h,k)$ is given as:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

Plug in $h=-4,k=-3,a=5,b=3$ and determine the equation.

$\frac{(x-(-4))^2}{5^2}+\frac{(y-(-3))^2}{3^2}=1\\\\\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Therefore, the equation of the ellipse is:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

6 0

3 years ago