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3 years ago

12

★PLSSSSS HELPPPPPPPPP★

1 answer:

SashulF [63]3 years ago

5 0

9514 1404 393

Answer:

(-2, 3)

Step-by-step explanation:

Of the choices offered, the only point that is on the line is (-2, 3).

You might be interested in

What is the inverse of this function? (Function in photo)

valkas [14]

Given:

The function is

$f(x)=-\dfrac{1}{2}\sqrt{x+3},x\geq -3$

To find:

The inverse of the given function.

Solution:

We have,

$f(x)=-\dfrac{1}{2}\sqrt{x+3}$

Put f(x)=y.

$y=-\dfrac{1}{2}\sqrt{x+3}$

Interchange x and y.

$x=-\dfrac{1}{2}\sqrt{y+3}$

Isolate the variable y.

$-2x=\sqrt{y+3}$

$(-2x)^2=y+3$

$4x^2-3=y$

$y=4x^2-3$

Putting $y=f^{-1}(x)$ , we get

$f^{-1}(x)=4x^2-3$

For $x\geq -3$ ,

$x+3\geq 0$

$\sqrt{x+3}\geq 0$

$-\dfrac{1}{2}\sqrt{x+3}\leq 0$

$f(x)\leq 0$

It means for function $f(x)\leq 0$ and for inverse function $x\leq 0$ because the range of the function is the domain of inverse function.

Therefore, $f^{-1}(x)=4x^2-3$ for $x\leq 0$ .

3 0

3 years ago

3 (m - 1) = 5m + 3 - 2m just solve it plz<br><br><br><br> dont answer if u dont know >:(

FinnZ [79.3K]

Answer:

No solution

Step-by-step explanation:

To solve the equation, we would have to find the value of m that would make the equality true.

3(m - 1) = 5m + 3 - 2m

Distribute the 3 on the left side.

3m - 3 = 5m + 3 - 2m

Combine like terms on right side.

3m - 3 = 3m + 3

Add 3 to both sides.

3m = 3m + 6

Subtract 3m from both sides.

0 = 6

Since 0 does not equal 6, there would be no solution to the equation, since no value of m would make 0 = 6 true.

I hope you find my answer and explanation to be helpful. Happy studying.

6 0

3 years ago

Read 2 more answers

Hey guys can help me please

ZanzabumX [31]

Answer:

Can you select multiple answers to this question? then option A, B and C all three applies, if only one the go for option C, since that's the major change happens to the parent function

8 0

3 years ago

Find the distance between lines 8x−15y+5=0 and 16x−30y−12=0.

AlladinOne [14]

Answer:

The distance between the two parallel lines is 11/17 units

Step-by-step explanation:

we have

8x-15y+5=0 -----> equation A

16x-30y-12=0 ----> equation B

Divide by 2 both sides equation B

8x-15y-6=0 ----> equation C

Compare equation A and equation C

Line A and Line C are parallel lines with different y-intercept

step 1

Find the slope of the parallel lines (The slope of two parallel lines is the same)

8x-15y+5=0

15y=8x+5

$y=\frac{8}{15}x+\frac{1}{3}$

the slope is

$m=\frac{8}{15}$

step 2

Find the slope of a line perpendicular to the given lines

Remember that

If two lines are perpendicular then their slopes are opposite reciprocal (the product of their slopes is -1)

$m1*m2=-1$

we have

$m1=\frac{8}{15}$

therefore

$m2=-\frac{15}{8}$

step 3

Find the equation of the line perpendicular to the given lines

assume any point that lie on line A

$y=\frac{8}{15}x+\frac{1}{3}$

For $x=0$

$y=\frac{1}{3}$

To find the equation of the line we have

$point\ (0,1/3)$ ---> is the y-intercept

$m=-\frac{15}{8}$

The equation in slope intercept form is

$y=-\frac{15}{8}x+\frac{1}{3}$ -----> equation D

step 4

Find the intersection point of the perpendicular line with the Line C

we have the system of equations

$y=-\frac{15}{8}x+\frac{1}{3}$ ----> equation D

$8x-15y-6=0$ ----> $y=\frac{8}{15}x-\frac{2}{5}$ ----> equation E

equate equation D and equation E and solve for x

$\frac{8}{15}x-\frac{2}{5}=-\frac{15}{8}x+\frac{1}{3}$

$\frac{8}{15}x+\frac{15}{8}x=\frac{1}{3}+\frac{2}{5}$

Multiply by 120 both sides to remove fractions

$64x+225x=40+48$

$289x=88$

$x=88/289$

Find the value of y

$y=-\frac{15}{8}(88/289)+\frac{1}{3}$

$y=-\frac{206}{867}$

the intersection point is $(\frac{88}{289},-\frac{206}{867})$

step 5

Find the distance between the points $(0,\frac{1}{3})$ and $(\frac{88}{289},-\frac{206}{867})$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$d=\sqrt{(-\frac{206}{867}-\frac{1}{3})^{2}+(\frac{88}{289}-0)^{2}}$

$d=\sqrt{(-\frac{495}{867})^{2}+(\frac{88}{289})^{2}}$

$d=\sqrt{(\frac{245,025}{751,689})+(\frac{7,744}{83,521})}$

$d=\sqrt{\frac{314,721}{751,689}}$

$d=\frac{561}{867}\ units$

Simplify

$d=\frac{11}{17}\ units$

therefore

The distance between the two parallel lines is 11/17 units

see the attached figure to better understand the problem

4 0

3 years ago

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image. E(−4, 1), F(−3, 3), G(−2,

MA_775_DIABLO [31]

Answer:

E'(4, -1), F'(3, -3), G'(2, -3), H'(1, -1)

Step-by-step explanation:

Rotation 180° about the origin negates each coordinate value.

6 0

3 years ago