Ask question

Ask question

All categories

2 years ago

6

Write the equation of a line that passes through the point (-2,7) and perpendicular to a line that passes through the points (-6

,1) and (0,4)

1 answer:

Alexandra [31]2 years ago

8 0

Answer:

Step-by-step explanation:

(x₁, y₁)= (-6,1) and (x₂,y₂) = (0,4)

Slope = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$= \dfrac{4-1}{0-[-6]}\\\\=\dfrac{3}{0+6}\\\\=\dfrac{3}{6}\\\\=\dfrac{1}{2}$

Slope of line perpendicular to it = -1/m

$= \dfrac{-1}{\dfrac{1}{2}}\\\\=-1*\dfrac{2}{1}=-2$

Equation : y =mx +b

y = -2x + b

Plugin x = -2 and y =7 in the above equation

7 = -2*(-2) + b

7 = 4 + b

7-4=b

3 = b

Equation: y =-2x + 3

You might be interested in

You want to build supports at each end of a table in the shape of a triangle. What type of triangle would you use to act as the

german

right angle because its always perpendicular to the floor

8 0

2 years ago

I need help!!!!!!!!!!!!

qaws [65]

When I do what the problem statement says, I get 47° for the left angle and 58° for the right one. They are not congruent.

4 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

Quirky Elf has 3.36 kilograms of fudge and 2.28 kilograms of divinity candy. He estimates that he has about 1 more kilogram of f

inysia [295]

Answer:

yes

Step-by-step explanation:

7 0

2 years ago

Two opposite sides of a rectangle are each of length x. If the perimeter of the rectangle is 12, then the area, as a function of

Reil [10]

Answer: $x(6-x)$

Step-by-step explanation:

Given : Two opposite sides of a rectangle are each of length x.

Let the other adjacent side be y.

The perimeter of the rectangle is 12 units.

Perimeter of rectangle is given by :-

$P=2(\text{Sum of adjacent sides})\\\\\Rightarrow\ 12=2(x+y)\\\\\Rightarrow\ x+y=\dfrac{12}{2}=6\\\\\Rightarriow\ y=6-x$

The area of rectangle is given by :-

$A=\text{product of two adjacent sides}\\\\\Rightarrow\ A=xy=x(6-x)$

Hence, the area as a function x = $x(6-x)$

7 0

3 years ago

Read 2 more answers