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2 years ago

PLSSS HELP DUE SOONNNN ILL GIVE BRAINLIEST IF ALL QUESTIONS R ANSWERED

Mathematics

1 answer:

Alexeev081 [22]2 years ago

3 0

Answer:

9. y=9h+4

10. y=5r+10

11. y=4x+8

12. y=4n+9

Step-by-step explanation:

9. He gets 9 dollars per hour (h), and 4 dollars extra for gas.

So 9 dollars per hour plus 4 dollars for the gas

y=9h+4

10. Roses (r) are selling for 5 dollars each and there is a delivery fee of 10 dollars.

So 5 dollars per rose plus 10 dollar fee

y=5r+10

11. The tree increases by 4 feet per year (x), and it is already 8 feet tall.

So 4 feet per year plus 8 feet

y=4x+8

12. Each trophy (n) weighs 4 kilograms, and the box itself weighs 9 kilograms.

So 4 kilograms per trophy plus 9 kilograms for the box.

y=4n+9

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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