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4 years ago

How do i find the slope and y intercept ?

Mathematics

2 answers:

Svetach [21]4 years ago

8 0

Here are the instructions. Then we'll do your specific problem.

Step 1: Solve the equation for y, and write it in the form y = mx + b.

Step 2: The slope is m, and the y-intercept is b.

Now let's start.

-2x - 5y = -20

To solve for y, we first add 2x to both sides.

-5y = 2x - 20

Now we divide both sides by -5.

$y = -\dfrac{2}{5}x + 4$

Now we compare what we got with the form, y = mx + b.

We see that m = -2/5, and b = 4. m is the slope, and b is the y-intercept.

slope: -2/5

y-intercept: 4

Vika [28.1K]4 years ago

4 0

y=mx+B is the formula for finding the slop intercept

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Sin t sin 3t sin 5t = 1/4(-sin t + sin 3t +sin 7t - sin 9t).

Pavlova-9 [17]

Recall that

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

cos(a - b) = cos(a) cos(b) + sin(a) sin(b)

and by subtracting the first equation from the second, we get

cos(a - b) - cos(a + b) = 2 sin(a) sin(b)

So, we can write

sin(t) sin(3t) = 1/2 (cos(2t) - cos(4t))

and expanding the left side in the original equation gives

sin(t) sin(3t) sin(5t) = 1/2 (cos(2t) - cos(4t)) sin(5t)

… = 1/2 cos(2t) sin(5t) - 1/2 cos(4t) sin(5t)

Similarly, recall that

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

sin(a - b) = sin(a) cos(b) - cos(a) sin(b)

===> sin(a + b) + sin(a - b) = 2 sin(a) cos(b)

Then

cos(2t) sin(5t) = 1/2 (sin(7t) + sin(3t))

and

cos(4t) sin(5t) = 1/2 (sin(9t) + sin(t))

So we have

sin(t) sin(3t) sin(5t) = 1/2 cos(2t) sin(5t) - 1/2 cos(4t) sin(5t)

… = 1/2 (1/2 (sin(7t) + sin(3t))) - 1/2 (1/2 (sin(9t) + sin(t)))

… = 1/4 sin(7t) + 1/4 sin(3t) - 1/4 sin(9t) - 1/4 sin(t)

… = 1/4 (-sin(t) + sin(3t) + sin(7t) - sin(9t))

as required.

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