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scoray [572]
3 years ago
11

What is the range of this function?

Mathematics
2 answers:
xeze [42]3 years ago
7 0

Answer:

<h3>{-8, -3 , 5 , 7}</h3>

Step-by-step explanation:

<h2>About domain and Range:</h2>

Let R be a relation from A to B. Then the set of first components or the set of elements of A are called domain and the set of second components or the set of elements of B are called the range.

Hope this helps...

Good luck on your assignment..

melisa1 [442]3 years ago
3 0

The range will be the set of all the outputs.

So that's going to be {-8, -3, 5, 7}.

You might be interested in
Find two consecutive even positive integers whose product is 624.
dimaraw [331]
Let x=first even number then x+2 would be the next even number. (x)(x+2)=624. x^2 +2x=624. x^2 +2x -624=0. Factor or use the quadratic formula. (x+26)(x-24)=0 x=-26 or x=24 We want the positive solution so x=24 and x+2=26 24*26=624 so it works.
A square root can start us ...
625 = 25 x 25 ==> 24 x 26 will do it!
4 0
4 years ago
Plz help me with this math ​
erma4kov [3.2K]

Answer:

$68.13

Step-by-step explanation:

$7.29x3=$21.87

$5x2=$10.00

$21.87 + $10.00=$31.87

$100.00-$31.87=$68.13

5 0
4 years ago
Write the equation of a circle with a center at<br> (2, -9) and a radius of 7<br> Helpppp
siniylev [52]

Answer:

x^2+y^2+14y+36=0

Step-by-step explanation:

Given that,

Center = (2,-9)

Radius of the circle, r = 7

We need to find the equation of a circle with a center at  (2, -9) and a radius of 7. It can be given by :

(x-h)^2+(y-k)^2=r^2

Put h = 2, k = -9 and r = 7 So,

(x-2)^2+(y-(-9))^2=7^2\\\\x^2+4-4x+y^2+81+18y=49\\\\x^2+y^2+14y+85-49=0\\\\x^2+y^2+14y+36=0

Hence, this is the required solution.

5 0
3 years ago
The type of flower shown in the table make up an arrangement what percent of the flowers in the range that are roses
sdas [7]
Can you show the table please?
4 0
4 years ago
Let m(x) = x / x−1
allsm [11]

Answer with Step-by-step explanation:

We are given that a function

m(x)=\frac{x}{x-1}

a.We have to find the inverse of m.

Suppose, y=m(x)=\frac{x}{x-1}

yx-y=x

yx-x=y

x(y-1)=y

x=\frac{y}{y-1}

Replace x by y and y by x

y=\frac{x}{x-1}

Substitute y=m^{-1}(x)

g(x)=m^{-1}(x)=\frac{x}{x-1}

b.When the inverse function of given function is \frac{x}{x-1}

Then , we get fog(x)=f(g(x))=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}

fog(x)=\frac{x}{x-x+1}=x=I(x)

When f and g are inverse to each other then fog(x)=Identity function.

c.If f(x)=x

It is self inverse function.

8 0
4 years ago
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