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OleMash [197]
4 years ago
11

Find the domain and range for the absolute value function f(x)=| x+2 |-3

Mathematics
1 answer:
vovangra [49]4 years ago
4 0
F(x) = |x + 2| - 3
   0 = |x + 2| - 3
   0 = x + 2 - 3
   0 = x - 1
+ 1      + 1
    1 = x

Domain: x > 1 and x < 1
Solution Set of the Domain: {x| 1 > x > 1} and {x|x 1 < x < 1} or (1, 1)

f(x) = |x + 2| - 3
f(x) = |0 + 2| - 3
f(x) = |2| - 3
f(x) = 2 - 3
f(x) = -1

Range: x ≤ 3
Solution Set of the Range: {x|x ≤ 3} or (-∞, 3]
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A radioactive substance decays exponentially. A scientist begins with 190 milligrams of a radioactive substance. After 19 hours,
Lelu [443]

Answer:

  51 milligrams

Step-by-step explanation:

Exponential growth or decay can be modeled by the equation ...

  y = a·b^(x/c)

where 'a' is the initial value, 'b' is the "growth factor", and 'c' is the time period over which that growth factor applies. The time period units for 'c' and x need to be the same.

In this problem, we're told the initial value is a = 190 mg, and the value decays to 95 mg in 19 hours. This tells us the "growth factor" is ...

  b = 95/190 = 1/2

  c = 19 hours

Then, for x in hours the remaining amount can be modeled by ...

  y = 190·(1/2)^(x/19)

__

After 36 hours, we have x=36, so the remaining amount is ...

  y = 190·(1/2)^(36/19) ≈ 51.095 . . . . milligrams

About 51 mg will remain after 36 hours.

8 0
2 years ago
Mya and 3 of her friends went out to eat, the total bill was 45.99 they gave her a 20% tip and then split the bill. How much did
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3 years ago
If The 3rd term of an A.S is and 5th term is . What is an A.P?​
Bad White [126]

Question:

If The 3rd term of an A.S  23 is and 5th term is 15 .P What is an A.?​

Answer:

A.P is   31, 27, 23 ,19, 15, 11, 7, 3

Step-by-step explanation:

Given:

3rd term =  23

5th term =  15

To Find:

The arithmetic Progression = ?

Solution:

Let

a_1 be the first term and d be difference

Then the arithmetic progression will be ,

a_1 , (a_1+d),(a_1 +2d),......(a_1+(n-1)d)

From the question ,

third term = 13

(a_1 +2d) = 23-------------(1)

Similarly

(a_1 +4d) = 15-----------------(2)

Subtracting (1) from  (2) we get

-2d =8

d = \frac{8}{2}

d = -4

Substituting d = 4 in equation 1, we get

(a_1 +2(-4)) = 23

(a_1 -8) = 23

a_1 = 23 +8

a_1 = 31

a_1 , (a_1+d),(a_1 +2d),......(a_1+(n-1)d)

=  31 ,(31 +(-4)) ,(31 +2(-4)), (31+3(-4)).........

=>  31, 27, 23 , 19, 15, 11,7,3

3 0
3 years ago
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