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liq [111]
3 years ago
14

How do you find the inverse of a function

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
You use the reciprocal slope and negative coefficient. 
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Simplify completely the quantity 3 times x to the 4th power plus 5 times x to the 2nd power plus 2 times x all over x. (2 points
BartSMP [9]
3x^3+5x+2 is the answer
5 0
3 years ago
Read 2 more answers
Determine whether the relation is a function: {(6, 1), (8, –3), (6, 7)}.
ICE Princess25 [194]

Hello there! The answer would be the first one, or No. At least one output results in two inputs.

When dealing with functions: you must remember that x does not repeat. So, lets look at the relation given, {(6, 1), (8, –3), (6, 7)}. You can see that x does repeat, so this is not a function. This eliminates second and fourth option choices. Out of the options A and C, A would be your choice since x is the input value and y is the output value, and there are two input values.

Hope his helps and have a  great day!

5 0
3 years ago
Graph the function represented in the table on the coordinate plane.
Umnica [9.8K]
First find out the function for the given table.
Find the slope.
m= (y2-y1)/(x2-x1)
m=(-1-(-3)) / (-1-(-2))
m=(-1+3) / (-1+2)
m=2/1
m=2
Now equation of line:
y-y1=m(x-x1)
y-(-3)=2(x-(-2))
y+3=2(x+2)
y+3=2x+4
y=2x+1
Now plot these points on the graph, we get the attached graph.

7 0
3 years ago
Read 2 more answers
The average of A and B is 6. The average of X, Y, and Z is 16. What is the average of A, B, X, Y, and Z?
ser-zykov [4K]

Answer:

12

Step-by-step explanation:

Average of A and B is 6

(A+B) /2 = 6

Multiply each side by 2

A+B = 12

average of X, Y, and Z is 16

(X+Y+Z)/3 = 16

Multiply each side by 3

(X+Y+Z) = 48

To find the average of all of them

(A+B+X+Y+Z)/5

replace A+B with 12 and X+Y+Z with 48

(12+48)/5

60/5 = 12

4 0
3 years ago
URGENT!!<br> The difference between the roots of the quadratic equation x2−12x+q=0 is 2. Find q
Tasya [4]

Answer:

-24

Step-by-step explanation:

substitute by 2

2²-12×2+q=0

q=-24

8 0
3 years ago
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