How do you multiply a fraction by a whole number

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

Harrizon [31]

Answer:

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

Step-by-step explanation:

We need to Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

First we will find slope of the line given in graph

The slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have points (1,6) and (2,2)

Slope is:

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-6}{2-1}\\Slope=\frac{-4}{2}\\Slope=-2$

Part a)

Write an equation of the line that passes through the given point and is (a) parallel

When the lines are parallel, they have same slope. So, slope of required line: m = -2

Using slope m =-2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=-2(4)+b\\3=-8+b\\b=3+8\\b=11$

The equation of line will be:

$y=mx+b\\y=-2x+11$

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Part b)

Write an equation of the line that passes through the given point and is (b) perpendicular

When the lines are perpendicular they have opposite slope. So, slope of required line: m = 1/2

Using slope m =1/2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=\frac{1}{2}(4)+b\\3=2+b\\b=3-2\\b=1$

The equation of line will be:

$y=mx+b\\y=\frac{1}{2}x+1$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

6 0

3 years ago

(01.02 HC)

Phoenix [80]

By applying radical, power and rational properties we have the following results after simplifying expressions:

$x^{\frac{3}{4} }$
x
$x^{\frac{11}{10} }$
x

<h3>How to reduce radical and rational expressions into a single power expression with rational exponent</h3>

In this question we need to simplify four given expressions by using the following radical and power properties:

$\sqrt[n]{x} = x^{\frac{1}{n} }$ (1)
$x^{m+n} = x^{m}\cdot x^{n}$ (2)
$x^{m-n} = \frac{x^{m}}{x^{n}}$ (3)
$(x^{m})^{n} = x^{m\cdot n}$ (4)

Now we proceed to simplify each of the four expressions:

$\sqrt[4]{x^{3}} = x^{3/4}$
$\frac{1}{x^{-1}} = (x^{-1})^{-1} = x$
$\sqrt[10]{x^{5}\cdot x^{4}\cdot x^{2}} = \sqrt[10]{x^{11}} = x^{\frac{11}{10} }$
$x^{\frac{1}{3} }\cdot x^{\frac{1}{3} }\cdot x^{\frac{1}{3} } = x$

To learn more on power functions: brainly.com/question/18719083

#SPJ1

3 0

3 years ago

Express the angle in radian measure. 60°

Strike441 [17]

Pi radians =180 degrees:

That is, π rad = 180°

60° in radian measure = 60° × π / 180°

60° × π / 180° = 1/3 π rad
= 0.33333333333π rad
= 1.0471975512 rad

7 0

3 years ago

Can a system of two linear equations have exactly two solutions?

Elenna [48]

Most linear systems you will encounter will have exactly one solution. However, it is possible that there are no solutions, or infinitely many. (It is not possible that there are exactly two solutions.) The word unique in this context means there is a solution, and it's the only one.

Hope this helps.

3 0

3 years ago

Which ordered pair can be included in the table of values shown below for the equation p=25x-2?

pishuonlain [190]

Answer:

(0,-2)

Step-by-step explanation:

So since you are given the equation, p=25x-2, all you really need to do is plug in the ordered pairs and see which satisfies the equation.

This means makes both side have the same number value.

Let's try (x,p)=(-2,-2):

p=25x-2 with (x,p)=(-2,-2)

-2=25(-2)-2

-2=-50-2

-2=-52

-2 and -52 do not represent the same number so (-2,-2) is not a point on the given line.

Let's try (x,p)=(-2,0):

p=25 x -2 with (x,p)=(-2,0)

0=25(-2)-2

0=-50-2

0=-52

0 and -52 do not represent the same number so (-2,0) is not a point on the given line.

Let's try (x,p)=(0,-2):

p=25 x -2 with (x,p)=(0,-2)

-2=25(0)-2

-2=0 -2

-2=-2

-2 and -2 are the same number and so (0,-2) is a point on the line.

Let's try (x,p)=(0,0)

p=25 x -2 with (x,p)=(0,0)

0 =25(0)-2

0 = 0 -2

0 = -2

0 and -2 are not the same number so (0,0) is not a point on this line.

6 0

4 years ago

Read 2 more answers