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notka56 [123]
3 years ago
12

What is the range of the following function?

Mathematics
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

Option d

Step-by-step explanation:

Range of a function is the set of y-values for every x-values.

Domain is the set of all values of variable 'x'

From the picture attached,

y-values of the given function varies from negative infinity to positive infinity.

Therefore, range of the function graphed is a set of all real numbers.

Option d is the correct option.

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Simplify 24÷⌈32÷(8−5)⌉−(−6)
Andrews [41]
Decimal form - 8.24
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8 0
3 years ago
How do i calculate this? PV = 100,000(1+0.03)
meriva

Answer:

  103,000

Step-by-step explanation:

Use your knowledge of decimal arithmetic. Or, use a calculator.

  100,000(1 +.03) = 100,000·1.03 = 1000·103 = 103,000

5 0
3 years ago
What are the factors of -20 that have a sum of -1
qaws [65]

Answer:

The factors of -20 are:

1, 2, 4, 5, 10, 20

-1, -2, -4, -5, -10, -20

Step-by-step explanation:

6 0
3 years ago
Hi can anyone do this for me please do it on a piece of paper
Vsevolod [243]

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

We don't know the drivers' names or when or where they started. We have made the assumption that the second equation pertains to Kylie.

Each line is plotted with the appropriate slope and y-intercept. The slope is the coefficient of x, and represents the "rise" for each unit of "run" to the right.

8 0
3 years ago
Find S18 for geometric series given a5=-6 and a2= -48
Ganezh [65]

an = a1r^(n-1)

a5 = a1 r^(5-1)

-6 =a1 r^4


a2 = a1 r^(2-1)

-48 = a1 r


divide

-6 =a1 r^4

----------------    yields   1/8 = r^3      take the cube root  or each side

-48 = a1 r                     1/2 = r


an = a1r^(n-1)

an = a1 (1/2)^ (n-1)

-48 = a1 (1/2) ^1

divide by 1/2

-96 = a1


an = -96 (1/2)^ (n-1)


the sum

Sn = a1[(r^n - 1/(r - 1)]

S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]

       =-96 [ (1/2) ^ 17 -1 /-1/2]

      = 192 * [-131071/131072]

 approximately -192

     

       

     

3 0
4 years ago
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