Find the equation of the line that passes through (1,4) and is parallel to 3x+y+2=0. Leave your answer in the form y=mx+c.

Mathematics

1 answer:

Mila [183]2 years ago

3 0

Answer:

y = -3x + 7.

Step-by-step explanation:

First find the slope of the given line by converting to intercept form:

3x + y + 2 = 0

y = -3x - 2

So the slope is -3.

Then the line we want has a slope of -3 also (as it is parallel).

y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point on the line:

m = -3, x1 = 1 and y1 = 4, so we have:

y - 4 = -3(x - 1)

y = -3x + 3 + 4

y = -3x + 7.

You might be interested in

What is 3 2/5 times 2 1/3

Serjik [45]

$3 \frac{2}{5} \cdot2 \frac{1}{3} = \frac{3\cdot5+2}{5} \cdot \frac{2\cdot3+1}{3} = \frac{17}{5} \cdot \frac{7}{3} = \frac{17\cdot7}{5\cdot3} = \frac{119}{15} =7 \frac{14}{15}$

7 0

3 years ago

Read 2 more answers

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .