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beks73 [17]
3 years ago
15

Find a statement that is equivalent to the absolute value equation |9x−9|=7.

Mathematics
1 answer:
Olegator [25]3 years ago
8 0

Answer:9x-9=7 or 9x-9=-7

Step-by-step explanation:

An absolute value is the distance from zero, therefore the answer can not be negative. In this case, 9x-9=7 and 9x-9=-7 both work, because 7 and -7 are the same distance from zero.

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Please help me with these questions I am begging anyone please help
Y_Kistochka [10]

Answer:

1)rs.6000

Step-by-step explanation:

1)30 000*20/100=rs.6000

Sorry I don't know the answers for other questions cause I don't know outright price

5 0
3 years ago
Since Marnie has bought her car,the value has gone down 15%.If her car is 13000 right now,how much was it worth when she bought
Serggg [28]
Percentage of depreciation of the car that Marie bought = 15%
Present value of the car = 13000
Let us assume that the value of the car at the time of buying = x
So
The percentage valuation of the car now = (100 - 15) percent
                                                                 = 85 percent
Then
(85/100) * x = 13000
85x = 13000 * 100
85x = 1300000
x = 1300000/85
  = 15294.12
So the actual cost of the car is 15294.12. I hope the procedure is clear enough for you to understand.
6 0
3 years ago
Mary moves 3 steps backward as she dances​
IgorLugansk [536]
Where’s the question????
3 0
2 years ago
What is the radical form of the expression (2x^4y^5)^3/8
s344n2d4d5 [400]

We are given expression: (2x^4y^5)^{3/8}

Let us distribute 3/8 over exponents in parenthesis, we get

(2^{3/8}x^{4\times 3/8}y^{5\times 3/8}) = (2^{3/8}x^{12/8}y^{15/8})

= (2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} )

We can get x and y out of the radical because, we get whlole number 1 for x and y exponents for the mixed fractions.

So, we could write it as.

(2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} ) = xy(2^{\frac{3}{8} }x^{\frac{4}{8}} y^{\frac{7}{8}} )

Now, we could write inside expression of parenthesis in radical form.

xy\sqrt[8]{2x^{3}x^4y^7}

8 0
3 years ago
Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

C(t) = C(0)e^{rt}

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So C(0) = 110

Husband

Half life of 4 hours. So

C(4) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{4r}

e^{4r} = 0.5

Applying ln to both sides

\ln{e^{4r}} = \ln{0.5}

4r = \ln{0.5}

r = \frac{\ln{0.5}}{4}

r = -0.1733

So for the husband

C(t) = 110e^{-0.1733t}

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.1733t}

C(11) = 110e^{-0.1733*11} = 16.35

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

C(10) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{10}

e^{10r} = 0.5

Applying ln to both sides

\ln{e^{10r}} = \ln{0.5}

10r = \ln{0.5}

r = \frac{\ln{0.5}}{10}

r = -0.0693

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.0693t}

C(11) = 110e^{-0.0693*11} = 51.33

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0
3 years ago
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