Ask question

Ask question

All categories

DanielleElmas [232]

2 years ago

13

The equation of line AB is y = 2x + 4. Write an equation of a line parallel to line AB in slope-intercept form that contains poi

nt (3, −2). A. y = 2x + 4 B. y = negative 1 over 2 , x − 1 over 2 C. y = − 1 over 2 , x − 7 over 2 D. y = 2x − 8

1 answer:

ale4655 [162]2 years ago

7 0

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

y = 2x + 4

comparing with the above equation for a line

The slope / m = 2

Since the lines are parallel their slope are also the same

Slope of parallel line = 2

So we have

The equation of the line using point

( 3 , -2) and slope 2 is

y + 2 = 2( x - 3)

y + 2 = 2x - 6

y = 2x - 6 - 2

We have the final answer as

<h3>y = 2x - 8</h3>

Hope this helps you

You might be interested in

If you got it correct I will follow and mark it as brilliant answer

Finger [1]

A) 9 + 3x + 6 + 9x + 5
b) 12x + 20

6 0

2 years ago

Which of the following equations is equivalent to 3(2p + 4) – (3p – 4) = 43?

kupik [55]

I just solved it for you and the answer i got was D

4 0

2 years ago

45 POINTS AND BRAINLIEST IF CORRECT

natita [175]

Answer:

2x+100

Step-by-step explanation:

5 0

3 years ago

Choose the end behavior of the graph of each polynomial function

valentina_108 [34]

Answer:

D, A, B

Step-by-step explanation:

The sign of the leading coefficient always tells you the end behavior on the right: positive = rises; negative = falls.

The degree of the polynomial tells you how the left- and right-end behaviors compare: even = they are the same; odd = they are opposites.

__

A) negative leading coefficient, even degree:

falls to the left, falls to the right (D)

__

B) positive leading coefficient, odd degree:

falls to the left, rises to the right (A)

__

C) negative leading coefficient, odd degree

rises to the left, falls to the right (B)

5 0

2 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago