All categories

3 years ago

Find the equation of the line that is perpendicular to y=5 through the point (-4,-6)

Mathematics

1 answer:

Nikolay [14]3 years ago

4 0

Answer:

$x=-4$

Step-by-step explanation:

we know that

The given equation y=5 is a horizontal line (is parallel to the x-axis)

The slope of the given line is equal to zero

A perpendicular line to the given line is a vertical line (parallel to the y-axis)

The equation of a vertical line is equal to the x-coordinate of the point that passes through it

The point that passes through it is (-4,-6)

therefore

The equation of the perpendicular line is

$x=-4$

You might be interested in

Jimmie adds 3.24, 4.294 and 2.073 . How much more than 5.31 is the sum

Brrunno [24]

4.477 hope that helps.

8 0

3 years ago

Read 2 more answers

Which of the following is the function of f(x)?

Natasha2012 [34]

Answer:

f(x) = 8(x-3)

Step-by-step explanation:

F^ -1 ( x) = x/8 +3

Let y = x/8+3

To find the inverse

Exchange x and y

x = y/8+3

Solve for y

x-3 = y/8+3-3

x-3 = y/8

Multiply each side by 8

8(x-3) = y/8 * 8

8(x-3) = y

The inverse of the inverse is the function so

f(x) = 8(x-3)

5 0

3 years ago

Read 2 more answers

What is 50/80 in simplest form? Thanks for your awesome help people

Sliva [168]

The answer is 5/8 because they both are divisable by 10

7 0

3 years ago

sarah, natasha and richard share some sweets in the ratio 5:2:4. sarah gets 60 sweets. how many did richard get?

Sindrei [870]

Answer:

48 sweets

Step-by-step explanation:

5:2:4

Sarah: 5

Natasha:2

Richard:4

To find out how many sweets Richard got, we need to use 60 as the lcm. Since, we have a definite constant of 60 we need to divide it by the presiding ratio.

60/5 = 12

That means for a ratio of 1, everyone in the group recieves 12.

Now, we plug it in for each of the members.

5x = Sarah

2x = Natasha

4x = Richard

Sarah = 60 sweets

Natasha = 24 sweets

Richard = 48 sweets

6 0

3 years ago

Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (

Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

The correct equation is

$y+1=\sqrt{-2y-3}$

Solve for y

squared both sides

$(y+1)^2=(-2y-3)$

$(y^2+2y+1)=(-2y-3)$

$y^2+2y+1+2y+3=0$

$y^2+4y+4=0$

$(y+2)(y+2)=0$

$y=-2$

Verify

substitute the value of y in the original expression

$-2+1=\sqrt{-2(-2)-3}$

$-1=1$ ----> is not true

therefore

There are no true solutions to the equation.

4 0

3 years ago