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4 years ago

Write an equation of a line in slope intercept form that is perpendicular to the line 2x -3y = 12 and

Mathematics

1 answer:

ella [17]4 years ago

3 0

Answer:

The equation of the line would be y = -3/2x + 9

Step-by-step explanation:

In order to solve this, start by finding the slope of the original line. You can do this by solving for y.

2x - 3y = 12

-3y = -2x + 12

y = 2/3x - 4

Now that we have a slope of 2/3, we know that the perpendicular slope is -3/2 (since perpendicular lines have opposite and reciprocal slopes). We can use this and the new point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - 6 = -3/2(x - 2)

y - 6 = -3/2x + 3

y = -3/2x + 9

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HELP ME ASAP! Will give BRAINLIEST! Please read the question THEN answer correctly! No guessing. Check ALL that apply.

sdas [7]

Answer:

-2 & 1/3

Step-by-step explanation:

When you set the two equations equal to each other, they form to parallel, vertical lines. They cross the x-axis at -2 and 1/3. Hope this helps.

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3 years ago

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How many different arrangements are there of the letters a, b, c, d, e, f for which e is not last in line?

Olegator [25]

Hmm, there are 5 letters

if you cannot repeat the letters
hmm, let's find how many for 4 letters
there are 4! or 4*3*2*1=24
that's how many ways there are to arrange a,b,c, d
now we can multiply it by 4 because there are 4 choices for the last place (e is not included0
24*4=56
answer is 56 different arrangements

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3 years ago

Determine the best method to solve the system of equations. Then solve the system.

maxonik [38]

Answer:

x=1 y = -1/2

(1,-1/2)

Step-by-step explanation:

7x-2y=8

5x+2y=4

I would use elimination since we have 2y in one equation and -2y in the other

7x-2y=8

5x+2y=4

--------------------

12x = 12

Divide each side by 12

12x/12 = 12/12

x =1

The substitute back into equation 2

5(1) +2y = 4

5 +2y = 4

Subtract 5

5-5+2y = 4-5

2y = -1

Divide by 2

2y/2 = -1/2

y = -1/2

8 0

4 years ago

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What is the antiderivative of sin^2(x)cos^2(x)?

marin [14]

Answer:

$\frac{x}{8}-\frac{\sin(4x)}{32}+C$

Step-by-step explanation:

[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]

Recall that we can express the squared trig functions in terms of cos(2x). That is,

$\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.$

And so inverting these,

$\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)$ .

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

$\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).$

Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).

$\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))$

So then,

$\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).$

This is now integrable (phew),

$\int \sin^2x\cos^2x \ dx = \int \frac{1}{8}(1-\cos(4x)) \ dx = \frac{1}{8} \int (1-\cos(4x)) \ dx = \frac{1}{8}(x-\frac{1}{4}\sin(4x))+C.$

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