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Anton [14]
3 years ago
7

What are the domain and range of f(x)=(1/5)^x

Mathematics
1 answer:
shusha [124]3 years ago
6 0

Answer:

Infinite domain, no restrictions

Range: y \geq 0

Step-by-step explanation:

Using this graph attached (this is the graph of the function), we can determine the possible inputs (domain) and possible outputs (range). Notice how it almost reaches 0 but never does? Yeah, there's an infinite number of inputs.

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Please help solveeeeee
Juliette [100K]

Answer:

The answer is C and D.

Step-by-step explanation:

(-4)^1/2 = undefined

(-16)^1/4 = undefined

(-32)^1/5 = -2

(-8)^1/3 = -2

6 0
2 years ago
Find the surface area of the pyramid. Thanks you.
Mkey [24]

Answer:

85 ft^2

5*6*2=60

5*5=25

25+60=85

Step-by-step explanation:

6 0
3 years ago
!! Need help with the problem in the photo. !!
lianna [129]

Answer:

B 3x-2y=10

Step-by-step explanation:

This is correct because when solved it has the same slope as line <em>k.</em>

The slope of <em>k</em> is 3/2.

3x-2y=10

first subtract 3x from both sides

-2y=-3x+10

next divide both sides by -2

y=3/2x-5

this shows that the slope 3/2 is the same as line <em>k</em>

6 0
3 years ago
Read 2 more answers
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
3 years ago
Need help with this please help
elena-14-01-66 [18.8K]
There is 24 avocados in each crate

Divide 8640 and 360 to get 24

Hope this will help you!
7 0
2 years ago
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