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Write an equation that is perpendicular to the line 8x-4y=12 and pass through the origin

fenix001 [56]

Answers

Explanation

So first we’re gonna change the equation to y=Mx+b by subtracting the 8x and then dividing both sides by -4. This can be show like this

8x-4y=12
-8x. -8x

-4y=-8x+12
/-4. /-4

Y=2x-3

So now that we have the equation in Y=Mx+b. The problem still isn’t done because it’s asking us to set up an equation that’s perpendicular. Which means the slope becomes the opposite

So the slope isn’t 2 anymore but instead -1/2! So now that we know that we need to make the line go through the origin which is very easy because you can just set the y intercept to 0.

So now that the perpendicular line is

Y=-1/2x or y=-1/2x +0

I hope I explain this well!

Please mark as Brainliest

7 0

3 years ago

This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl

Lostsunrise [7]

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0

2 years ago

If a coin tossed 300 times .Then the nearest expected number for the tail to appear is

Eva8 [605]

Step-by-step explanation:

I think you can solve this question now

8 0

2 years ago

SOMEONE PLEASE HELP ME WITH THIS ASAP THIS IS ONE OF MY LAST QUESTIONS AND PLEASE SHOW WORK

-Dominant- [34]

Answer:

1296 ft^3.

Step-by-step explanation:

The ratio of the volumes = 1^3 : 6^3

= 1 : 216.

So volume of the larger figure = 6 * 216

= 1296 ft^3.

5 0

3 years ago

Please use the following images in order to answer the question correctly:

Valentin [98]

Given:

The line segments are AB, DB, AB, OB and BC

We need to determine the given line segments are radius, chord, diameter, secant or tangent of circle O.

Line segment AB:

A secant is a line segment that intersects the circle at two points.

Thus, the line segment AB is a secant.

Line segment DB ( $\overline{D B}$ ):

The diameter of the circle is line that passes through the center and touches the two ends of the circle.

Thus, from the figure, the line segment DB passes through the center and touches the two ends of the circle.

Hence, the line segment DB is the diameter.

Line segment AB ( $\overline{A B}$ ):

The line joining any two points on the circle is called the chord of the circle.

Thus, from the figure, the line segment AB touches the two points on the circle.

Hence, the line segment $\overline{A B}$ is the chord of the circle.

Line segment OB ( $\overline{O B}$ ):

The radius of the circle is any line from the center of the circle to any point on the circle.

Thus, from the figure, the line segment OB is a line from center of the circle to the point B on the circle.

Hence, the line segment $\overline{OB}$ is the radius of the circle.

Line segment BC:

A tangent is a point that touches the exterior point of the circle exactly one time.

Thus, from the figure, the line segment BC is a touches the circle exactly once.

Hence, the line segment BC is the tangent of the circle.

8 0

2 years ago

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