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iVinArrow [24]
3 years ago
14

Find the inverse function of f(x) = -1\3x+2

Mathematics
1 answer:
Helga [31]3 years ago
8 0
If f(x) = 2x – 5, find the inverse. This function passes the Horizontal Line Test which means it is a onetoone function that has aninverse. y = 2x – 5 Change f(x) to y. x = 2y – 5 Switch x and y. Solve for y by adding 5 to each side and then dividing each side by 2.
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48a2 + 32a2 + 16a by 4a?
lbvjy [14]

(48a^3 + 32a^2 + 16a) / 4a = ?

48a^3 / 4a = 12a^2

32a^2 / 4a = 8a

16a / 4a = 4

so

(48a^3 + 32a^2 + 16a) / 4a = 12a^2 + 8a + 4

Answer is D. 12a^2 + 8a + 4

7 0
3 years ago
Read 2 more answers
Which of these statements are correct about a line segment with a length of 12 units in a coordinate plane? Select all that appl
Vitek1552 [10]

Correct statements are:

If it is reflected across the y-axis, its length still will be 12 units.

If it is rotated 270° about the origin, its length still will be 12 units.

If it is translated 15 units up, its length still will be 12 units.

<u>Step-by-step explanation:</u>

Whatever it may be rotation, reflection or translation, the size of the line will never change. So length of the line is same as 12 units in the image.

So the wrong statements are

If its reflected across y = -x then the length will no longer be 12 units.

If it is rotated 90° about the origin, then the length will no longer be 12 units.

If it is translated 18 units to the right,  then the length will no longer be 12 units.

7 0
3 years ago
65 - 14 = 65-5-______
Fed [463]

Answer: 60? to be hostens i have no clue sorry i hope this help...

Step-by-step explanation:

7 0
3 years ago
Find the perimeter of the figure below notice that one side is not given
Anarel [89]
4+15+11+5+7+10= 50 cm

10cm  was the missing one
5 0
3 years ago
Read 2 more answers
What are the number of integral solutions of the equation 2x+2y+z=20 such that x&gt;=0 , y&gt;=0 , z&gt;=0?
zaharov [31]
2x+2y+z=20\\&#10;z=\dfrac{20}{2x+2y}\\&#10;z=\dfrac{10}{x+y}

Now, for z to be an integer, the sum x+y must be a divisor of 10. 
It has to be a positive divisor since z≥0. Also x≠y.

x+y=1 \\&#10;x=1-y\\&#10;
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.

x+y=2\\&#10;x=2-y\\
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.

x+y=5\\&#10;x=5-y
y can be 0,1,2,3,4 or 5 - 6 solutions

x+y=10\\&#10;x=10-y
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.

2+2+6+10=20 solutions in total.
6 0
3 years ago
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