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3 years ago

For the following exercises, find the equation of the line using given information (1, 7) and (3, 7)

Mathematics

1 answer:

arlik [135]3 years ago

6 0

y = 7

Horizontal line, therefore has a slope of zero

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Pls help meeee!!! I have benchmarks monday

quester [9]

Answer:

therefore the correct options are:

B. the equation is an exponential decay equation.

F. $28000 represents the initial cost of an automobile that depreciates 27% per year over the course of t years.

Step-by-step explanation:

i) the cost of an automobile is given by the equation is

C(t) = $28000\times(0.73)^{t}$ , where C(t) represents the cost and t represents the time in years

ii) the above equation is an exponential decay equation. It is a decay equation because growth rate is 0.73 which is less than 1 and greater than zero.

ii) when t = 0 then the cost is C(0) = 28000 represents the initial cost.

iii) from the equation in i) we can understand that the cost depreciates 27% per year over the course of t years.

therefore the correct options are:

B. the equation is an exponential decay equation.

F. $28000 represents the initial cost of an automobile that depreciates 27% per year over the course of t years.

5 0

3 years ago

Which functions have graphs that are less steep than the graph of f(x)= -4x^2

LiRa [457]

TTese are all parabolas whose axis of symmetry is the y axis. The negative coefficient means they open downwards and the positive mean they open upwards. The higher the absolute value of the coefficient is the greater the slope, so 2x^2 will be steeper than x^2 and -4x^2 will be steeper than - 2x^2.

So h(x) and j(x) are the answers.

8 0

2 years ago

The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

8 0

3 years ago

Read 2 more answers

PLEASE HELP ME I CANT FAIL THIS CLASS

ale4655 [162]

Answer:

It would be the first answer.

Step-by-step explanation:

y=-2x^2+4x-2

y=-2(-1)^2+4(-1)-2

y=-2(1)-4-2

y=-2-4-2

y=-6-2

y=-8

(-1,-8)

-Hope this helps!-

-Please mark as brainliest!-

-Good luck!-

4 0

3 years ago

Soren has 2 jobs. During the day, he works as an office clerk. In the evening, he works as a cashier. His office job pays him $1

goldenfox [79]

I'm pretty sure the answer is B) 11c+8.5(c-46)=470
hope i help :)

7 0

3 years ago

Read 2 more answers