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3 years ago

Write an equation in point slope and slope intercept form of a line that passes through the given point and

Mathematics

1 answer:

nekit [7.7K]3 years ago

7 0

Answer:

Point slope is ( Y+4) = 1/2(x+3)

Slope intercept is Y = 1/2(x) -5/2

Step-by-step explanation:

For the point slope form.

Given the point as (-3,-4)

And the gradient m = 1/2

Point slope form is

(Y - y1) = m(x-x1)

X1 = -3

Y1 = -4

(Y - y1) = m(x-x1)

(Y - (-4)) = 1/2(x -(-3))

( Y+4) = 1/2(x+3)

For the slopes intercept form

Y = mx + c

We can continue from where the point slope form stopped.

( Y+4) = 1/2(x+3)

2(y+4)= x+3

2y + 8 = x+3

2y = x+3-8

2y = x-5

Y = x/2 - 5/2

Y = 1/2(x) -5/2

Where -5/2 = c

1/2 = m

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Solve for y.<br> 8+4y= -8

Andrew [12]

Answer:

-4

Step-by-step explanation:

8+4y= -8

4y= -8 -8

4y= -16

divide both sides by the coeffient of the variable. in this case divide both sides by 4

y=-4

7 0

3 years ago

Read 2 more answers

Scheels has a kayak that regularly sells for $389. Every Friday in November it was marked down by 32%. What is the amount of dis

Y_Kistochka [10]

Answer:

1. $264.52 is the remaining 68%

2. Sales tax is $19.83

3. The total price is 283.83

Explanation:

1. 389 times 0.68 will get you an answer of $264.52

2. 264.52 times 0.075 is $19.83

3. Add your two answers of $264.52 and $19.83 and you'll get your answer of $283.83

8 0

3 years ago

A 2X2 square is centered at the origin. It is dilated by a factor of 3. What are coordinates of the vertices of the square?

Vesna [10]

Answer:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

Step-by-step explanation:

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1), B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

7 0

3 years ago

Which is the equation of a circle with its center at (-4,2) and a radius of 4?

Snowcat [4.5K]

Answer:

(x +4)^2 + (y-2)^2 = 16

Step-by-step explanation:

We can write the equation for a circle with the formula

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

We know the center is at (-4,2) and the radius is 4

(x- -4)^2 + (y-2)^2 = 4^2

(x +4)^2 + (y-2)^2 = 16

4 0

3 years ago

Solve 0 = (x – 4)2 – 1 by graphing the related function.

fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

$y=\left(x-\:4\right)^2-1$

Finding the x-intercept by substituting the value y = 0

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

$x=5$

$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

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3 years ago