All categories

3 years ago

8(2n - 5) = 3(6n - 2)

Mathematics

1 answer:

joja [24]3 years ago

5 0

Answer:

n=-17

Step-by-step explanation:

8(2n-5)=3(6n-2)

1) Distributive property

16x-40=18x-6

16x−40=18x−6

2) Subtract 16x16x from both sides.

-40=18x-6-16x

−40=18x−6−16x

3) Simplify 18x-6-16x18x−6−16x to 2x-62x−6.

-40=2x-6

−40=2x−6

4) Add 66 to both sides.

-40+6=2x

−40+6=2x

5) Simplify -40+6−40+6 to -34−34.

-34=2x

−34=2x

6) Divide both sides by 22.

-\frac{34}{2}=x

−

7) Simplify \frac{34}{2}

to 17

−17=x

8) Switch sides.

x=−17

You might be interested in

3(Ꮾ - f) -4 = 3f - 4

raketka [301]

3(6-f)-4=3f-4

First is expand

18-3f-4=3f-4

Combine like terms

14-3f=3f-4

18-3f=3f

18=6f

f=3

8 0

3 years ago

Read 2 more answers

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

4 years ago

Simplify the expression: 8(-3n + 8)

levacccp [35]

Answer:

-24n+64

Step-by-step explanation:

just distribute

7 0

3 years ago

Read 2 more answers

Calculate the average travel time for each distance, and then use the results to calculate. A 6-column table with 3 rows in the

BigorU [14]

The average travel time is simply the mean travel time between the given points

The average time that it takes for the car to travel the first 0.25m is 2.23s.
The average time to travel just between 0.25m and 0.50m is 0.9s.
Given the time taken to travel the second 0.25 m section, the velocity would be 0.28m/s.

(a) The average time to travel the first 0.25m

The time travel in the first 0.25m are: 2.24s, 2.21s and 2.23s.

So, the average time to travel is:

$Time = \frac{2.24s + 2.21s + 2.23s}{3}$