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Allushta [10]
3 years ago
15

What is the range of the function? f(x) = -2^x + 1

Mathematics
2 answers:
8090 [49]3 years ago
7 0

Answer:

all real numbers less than or equal to 0

Step-by-step explanation:

EDGE 2020

Readme [11.4K]3 years ago
6 0

{y|1 > y}, (-∞, 1); you did not specifically mention what notation you needed, so I wrote both [Internal Notation and Set Notation].

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Ricardo jogged up 864 steps in 13 1/2 minutes. What is Ricardo’s average number of steps per minute?
dangina [55]

Answer:

64 steps per minute

Step-by-step explanation:

864 / 13 1/2

    OR

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Both give us

64/1 = 64

3 0
3 years ago
A store is having a sale where movies or 15% off their original price a movie is on sale for $22.10. What was the original price
WINSTONCH [101]
........................................................
8 0
3 years ago
Please help and add explanation
Degger [83]

Answer:

y=-3x+10

Step-by-step explanation:

Since you know that the equation of the line is perpendicular to 1/3, you have to find the inverse of 1/3. You swap the denominator and numerator and make the number negative. In this case, it will be -3. Now that you know the slope is -3, you can use the equation y-y1=m(x-x1) where m is the slope. We know that the point (2,4) is in the graph so we can plug it into the equation. y-4=m(x-2). We know that the slope is -3 so we can plug it into the equation. y-4=-3(x-2). Simplify this to get y-4=-3x+6. Simplify this to get y=-3x+10. This means that the answer is y=-3x+10.

If this has helped you please mark as brainliest

3 0
3 years ago
21x + 862=? x= 4 I know how to do it but im too lazy to do it
LuckyWell [14K]

946

Hope this helps o(*^▽^*)┛

3 0
3 years ago
Solve for the variable X=5y for y
SSSSS [86.1K]

y= x/5. or x over 5. you are trying to isolate y. in other words you are trying to get y all by itself on one side of the equation. To do that, you need to divide 5 by both sides. It is important to do what you do on one side to the other because that keeps that values of the equation the same.
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4 years ago
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