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3 years ago

Solve the equation. -4(3 - 2x) + 2x = 2x - 8 A) x = 2 B) x = -1 C) x = 1/ 2 D) x = 1/ 3

2 answers:

7 0

-4(3-2x)+2x=2x-8
First Distribute your -4 to get -12+8x+2x=2x-8
Then combine like terms to get -12+10x=2x-8
Add 8 over to the -12 to get -4+10x=2x
Subtract 10x from 2x to get -4=-8x
Finally, divide -4 by -8x to get x=1/2.
The correct answer is Letter C.

vaieri [72.5K]3 years ago

4 0

-4(3-2x)+2x=2x-8
distribute -4 through (3-2x)= -12+8x
-12+8x+2x=2x-8
Cancel out 2x by subtracting it from both sides.
-12+8x=-8
Add 12 to both sides in order to get the -8x alone,
8x=4
Divide both sides by 8,
you will get 1/2 and that is C).

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4 years ago

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4 years ago

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3 years ago

Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the o

melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) $C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

c) $C(t)=-\frac{275}{14}t+\frac{288405}{7}$

d) t=2021

e) too early

Step-by-step explanation:

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915 concentration in year 2000

C(2014)=1915 concentration in year 2014

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

$m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}$

so:

$C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64$

therefore:

$C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

$y-y_{1}=m(x-x_{1})$

in this case:

$C(t)-C(t_1)=C'(t)(t-t_{1})$

so we get:

$C(t)-1915=-\frac{275}{14}(t-2000)$

and we can now solve for C(t) so we get:

$C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915$

for a final answer of:

$C(t)=-\frac{275}{14}t+\frac{288405}{7}$

So next we solve the equation for C(t)=1500 so we get:

$1500=-\frac{275}{14}t+\frac{288405}{7}$

$1500-\frac{288405}{7}=-\frac{275}{14}t$

$-\frac{277905}{7}=-\frac{275}{14}t$

$t=2021 \frac{7}{55}$

so we will reach that concentration level at the year 2021 approximately.

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

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3 years ago