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Anastasy [175]
3 years ago
6

A line passes through the points (1, 4) and (3, –4). Which is the equation of the line? y = negative 4 x + 8 y = –2x + 6 y = neg

ative one-foA line passes through the points (1, 4) and (3, –4). Which is the equation of the line? y = negative 4 x + 8 y = –2x + 6 y = negative one-fourth x + 2 y = 2x + 2urth x + 2 y = 2x + 2
Mathematics
2 answers:
Kruka [31]3 years ago
8 0

Answer:

y = -4x + 8

Step-by-step explanation:

The equation of a line is written in slope-intercept form : y = mx+b

m is the slope

We are given two points, so let's find the slope of the line first:

Slope = ΔY/ΔX = 4 - (-4) / 1 - 3 = 8 / -2 = -4

The slope is -4

So far, our equation is y = -4x + b

We can input a point's x and y value to find b, the y-intercept

Let's use point (1, 4)

4 = -4(1) + b

4 = -4 + b

b = 4 + 4

b = 8

The equation is y = -4x + 8

-Chetan K

kotykmax [81]3 years ago
4 0

Answer:

the answer is A

Step-by-step explanation:

y=-4x+8

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K is the midpoint of PQ, P has
Pachacha [2.7K]

Answer:

Coordinates of Q (x_2,y_2) \:are\: \mathbf{(7,16)}

Option D is correct option.

Step-by-step explanation:

We are given:

K is the midpoint of PQ

Coordinates of P = (-9,-4)

Coordinates of K = (-1,6)

We need to find coordinates of Q  (x_2,y_2)

We will use the formula of midpoint: Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

We are given midpoint K and x_1,y_1 the coordinates of P we need to find x_2,y_2 the coordinates of Q.

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(-1,6)=(\frac{-9+x_2}{2},\frac{-4+y_2}{2})\\

Now, we can write

-1=\frac{-9+x_2}{2}, 6=\frac{-4+y_2}{2}\\Simplifying:\\-2=-9+x_2\:,\: 12=-4+y_2\\-2+9=x_2\:,\: 12+4=+y_2\\x_2=7\:,\:y_2=16

So, we get coordinates of Q (x_2,y_2) \:are\: \mathbf{(7,16)}

Option D is correct option.

6 0
3 years ago
One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm
sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.  

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}

And now, 15 degrees is 11 times smaller than the other  

Then other angle = 11 times of 15 degrees

\text {Other angle }=11 \times 15=165 \text { degrees }

Now, sum of angles = 15 + 165 = 180 degrees.

As we expected their sum is 180 degrees. So the lines are parallel.

Hence, the given lines are parallel

5 0
3 years ago
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Answer:

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Step-by-step explanation:

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We know that pi = 3  

the diameter is 5 so the radius is 1/2 the diameter of 5/2 = 2.5

The height is 8

V = 1/3 ( 3) ( 2.5)^2 ( 8)

V = 50 in ^3

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Alecsey [184]

Answer:

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Step-by-step explanation:

88 X 56 = 4928 in2

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