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Vinvika [58]
3 years ago
7

Whats the answer to this??

Mathematics
1 answer:
nexus9112 [7]3 years ago
7 0
Same as before, apples with apples: (A), girls both in the numerator, total both in the denominator, righty?
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Find inverse of f(x)=x^3-9<br>Not sure of my answer!​
Katyanochek1 [597]

as you may already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".

\bf \stackrel{f(x)}{y}=x^3-9\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\underline{y}^3-9}\implies x+9=y^3\implies \sqrt[3]{x+9}=\stackrel{f^{-1}(x)}{y}

7 0
3 years ago
The vertex of the graph of f(x) = |x - 3| +6 is located at<br> Done<br> Intro
Igoryamba

Answer:

The vertex of the graph is located at the point (3,6)

Step-by-step explanation:

Here, we want to know where the vertex of the equation if plotted will be

To get this, what we have to do is to equate the expression that we have in the absolute value to zero

After this, we then proceed to solve for the value of x

We have this as;

x -3 = 0

hence;

x = 0 + 3

x = 3

to get the y-value of the vertex, we look at the value at the side of the absolute value

This value is 6 and thus, the y-value of the vertex point is 6

So the coordinates of the vertex is (3,6)

8 0
3 years ago
What is the volume of this figure?
Sav [38]

Answer:

90 cubic inches

Step-by-step explanation:

4 0
3 years ago
Write an equation of a line that is parallel to the line whose equation is 3y=x+6 and that passes through the point (-3,4).
Monica [59]
The standard form of this equation is

y=\frac{1}{3}x+2

The slope of two parallel lines i the same, so the slope of the line we're looking for is also \frac{1}{3}

Since the line passes through (-3,4), the equation is:

(y-4)=\frac{1}{3}(x+3)

\boxed{y=\frac{1}{3}x+5}





3 0
3 years ago
System of Equations <br> Need help<br> Trig
Yanka [14]

Looks like the system is

\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}

We can eliminate y by taking

(3x+2y+5z)-2(-2x+y+6z)=16-2(1)

\implies3x+2y+5z+4x-2y-12z=16-2

\implies7x-7z=14

\implies x-z=2

so that z=x-2, and

(7x+3y-4z)-3(-2x+y+6z)=11-3(1)

\implies7x+3y-4z+6x-3y-18z=11-3

\implies13x-22z=8

Substitute z=x-2 into this last equation and solve for x:

13x-22(x-2)=8

\implies13x-22x+44=8

\implies-9x=-36

\implies x=4

Then

z=x-2

\implies z=4-2

\implies z=2

Plug these values into any one of the original equation to solve for y:

-2x+y+6z=1

\implies-2(4)+y+6(2)=1

\implies-8+y+12=1

\implies y=-3

Hence the solution is x = 4, y = -3, and z = 2.

6 0
3 years ago
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