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3 years ago

What is the equation of the following line? Be sure to scroll down first to see

Mathematics

2 answers:

andriy [413]3 years ago

6 0

Answer:

y = 6x + 0

Step-by-step explanation:

Equation of a line

y = mx + c

Given

( 0 , 0) ( -1/2 , -3)

find the slope m

m = y2 - y1 / x2 - x1

x1 = 0

y1 = 0

x2 = -1/2

y2 = -3

Insert the values

m = y2 - y1 / x2 - x1

m = -3 - 0 / -1/2 - 0

= -3/-1/2

Minus cancels minus

= 3/1/2

= 3/1 ÷ 1/2

= 3/1 × 2/1

= 6/1

= 6

m = 6

Substitute any of the two points given into the equation of a line

y = mx + c

Where

y - intercept point y

x - intercept point x

m - slope of the line

c - intercept

(-1/2 , -3)

x = -1/2

y = -3

-3 = 6(-1/2) + c

-3 = -6/2 + c

-3 = -3 + c

-3 + 3 = c

c = 0

y = 6x + 0

The equation of the line is

y = 6x + 0

valentinak56 [21]3 years ago

5 0

Answer:

y= -6x on apex :)

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2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

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2 years ago