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bija089 [108]
3 years ago
11

Y=4x-6 and passes through point 8,12

Mathematics
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

Slope of required line if both lines are parallel: y=4x-20

Slope of required line if both lines are perpendicular: y=-1/4x+14

Step-by-step explanation:

We need to find equation of line while we are given another line having equation y=4x-6 and passes through point (8,12)

<u><em>Note: Since it is not given if the required line is parallel or perpendicular to the given line. I will solve for both cases.</em></u>

If Both lines are parallel the equation of new line will be:

We need to find slope and y-intercept to write equation of new line.

When lines are parallel there slopes are same

So, slope of given line y=4x-6 is 4 (Compare with general equation y=mx+b, m is slope so m=4)

Slope of new line is: m=4

Now finding y-intercept

Using slope m=4 and point (8,12) we can find y-intercept

y=mx+b\\12=4(8)+b\\12=32+b\\b=12-32\\b=-20

y-intercept of new line is: b=-20

Equation of required line having slope m=4 and y-intercept b=-20 is

y=mx+b\\y=4x-20

If Both lines are perpendicular the equation of new line will be:

We need to find slope and y-intercept to write equation of new line.

When lines are perpendicular there slopes are opposite of each other

So, slope of given line y=4x-6 is 4 (Compare with general equation y=mx+b, m is slope so m=4)

Slope of new line is: m=-1/4

Now finding y-intercept

Using slope m=-1/4 and point (8,12) we can find y-intercept

y=mx+b\\12=-\frac{1}{4} (8)+b\\12=-2+b\\b=12+2\\b=14

y-intercept of new line is: b=14

Equation of required line having slope m=-1/4 and y-intercept b=14 is

y=mx+b\\y=-\frac{1}{4} x+14

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Answer:C = 8

Answer:C = 8Step-by-step explanation:

-2(4c+16)^1/4+2=-2

Move the constant to the right and change the sign

-2(4c+16)^1/4= -2-2

transform expression

- 2  \sqrt[4]{4c - 16}  =  - 4

Divide each side by -2

\sqrt[4]{4c - 16 }  = 2

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5 0
3 years ago
Let (8,−3) be a point on the terminal side of θ. Find the exact values of cosθ, cscθ, and tanθ.
denis23 [38]

Answer:

\text{Cos}\theta=\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{x}{R}=\frac{8}{\sqrt{73}}

\text{Csc}\theta=-\frac{\sqrt{73}}{3}

\text{tan}\theta =\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{y}{x}=\frac{-3}{8}

Step-by-step explanation:

From the picture attached,

(8, -3) is a point on the terminal side of angle θ.

Therefore, distance 'R' from the origin will be,

R = \sqrt{x^{2}+y^{2}}

R = \sqrt{8^{2}+(-3)^2}

  = \sqrt{64+9}

  = \sqrt{73}

Therefore, Cosθ = \frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{x}{R}=\frac{8}{\sqrt{73}}

Sinθ = \frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{y}{R}=\frac{-3}{\sqrt{73} }

tanθ = \frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{y}{x}=\frac{-3}{8}

Cscθ = \frac{1}{\text{Sin}\theta}=\frac{R}{y}=-\frac{\sqrt{73}}{3}

6 0
4 years ago
Please solve if you know no trolls please im desparate
IgorC [24]
The question isn’t equal to 90° it’s equal to 180° because the line is straight it’s not a right angle. so when u equal the whole equation to 180°, you get x=18.75. then plug in the number to find the degree of each.
so angle of department would be 139.25° and angle of gas station would be 40.75°
3 0
3 years ago
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