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sdas [7]
2 years ago
7

If x is a number that is 7 more than y, how do you express x as a function of y?

Mathematics
1 answer:
Tomtit [17]2 years ago
8 0

Answer:

y = x - 7

Step-by-step explanation:

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Find the range of the following piecewise function.
mash [69]

Answer:

\huge\boxed{[-5;\ -2)\ \cup\ [3;\ 13)}

Step-by-step explanation:

The following piecewise functions are linear functions. The graph of any of them is a line segment.

We just need to calculate the value of the function at each end specified in the brace.

y=3x-2\ \text{if}\ -1\leq x

Substitute x =-1 and x = 0:

x=-1\\y=3(-1)-2=-3-2=-5\\\\x=0\\y=3(0)-2=0-2=-2

Range of this piece is [-5; -2)

y=2x+3\ \text{if}\ 0\leq x

Substitute x =0and x = 5:

x=0\\y=2(0)+3=0+3=3\\\\x=5\\y=2(5)+3=10+3=13

Range of this piece is [3; 13)

Therefore the range of the following piecewise function is:

[-5;\ -2)\ \cup\ [3;\ 13)

Look at the picture.

6 0
3 years ago
7(x+1)=21 solve for x
ElenaW [278]
7(x + 1) = 21
7x + 7 = 21
7x = 14
x = 2
5 0
3 years ago
5/6 divided by 1/2 simplest from
ss7ja [257]

Answer:

5/3

Step-by-step explanation:

(5/6)÷(1/2)

= (5/6) x 2

= 10/6 = 5/3

Hope this helps!

3 0
3 years ago
Read 2 more answers
The function f(x)= 3(1.4)' gives the length (in inches) of an image after being enlarged by 10% x times.
kondaur [170]

Answer:

3.99 inches

Step-by-step explanation:

<u><em>The correct question is</em></u>

The function f(x) = 3(1.1)^x gives the length (in inches) of an image after being enlarged by 10% x times.

What is the length of the image after it has been enlarged 3 times? Round your answer to the nearest  hundredth

we have

f(x)=3(1.1)^x

This is a exponential growth function

where

f(x) represent the length (in inches) of an image

x is the number of times the image is enlarged

so

For x=3

substitute in the exponential equation

f(x)=3(1.1)^3=3.99\ in

4 0
3 years ago
Tommy is training for an upcoming race at the track. Last Saturday, he ran 24 laps
Nezavi [6.7K]
Tommy ran 7 miles today for his track meet
7 0
4 years ago
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