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2 years ago

(6.) 6x – 4(-2x + 7) + 8 = 8x-19

Mathematics

2 answers:

Fed [463]2 years ago

7 0

Answer:

$\frac{1}{6}$

Step-by-step explanation:

1. you distrubute the -4 to the -2x and +7.

2.you combine like terms.

3.you get X in one side of the equation.

4.You divide 1 by 6.

marissa [1.9K]2 years ago

5 0

Answer:

The required value of x will be 1/6

Step-by-step explanation:

We have been given that an expression that is 6x – 4(-2x + 7) + 8 =8x-19.

We have to find the value of x.

First of all, We will distribute the terms in simplest form.

6x – 4(-2x + 7) + 8 =8x-19

6x + 8x - 28 + 8 = 8x - 19

We will combine the like terms.

6x + 8x -20 = 8x - 19

6x + 8x - 8x = -19 + 20

6x + 0 = 1

6x + 0 = 1 x = 1/6

Therefore, the required value of x will be 1/6.

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10 months ago

Read 2 more answers

In June, you bought 14.5 gallons of gasoline for $3.18 per gallon. The following February, you bought the same amount of gasolin

Nata [24]

Answer:

14.5 × 3.18 - 14.5 x 1.88

You paid $18.85 more in june than in February

Step-by-step explanation:

In order to find the answer to this question you will need to find how much you paid in June and how much you paid in February.

ex.14.5 x 3.18 - 14.5 x 1.88

To solve this you need to find the product of 14.5 x 3.18 and 14.5 x 1.88 and then subtract them from eachother.

ex.14.5 x 3.18 =46.11

14.5 x 1.88=27.26

46.11-27.26=18.85

3 0

2 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

2 years ago