Solve for x 6-12x = -54

Simplify: sec(θ) sin(θ) cot(θ)

krek1111 [17]

We have to simplify

sec(θ) sin(θ) cot(θ)

Now first of all let's simplify these separately , using reciprocal identities.

Sec(θ) = 1/cos(θ)

Sin(θ) is already simplified

Cot(θ)= cos(θ) / sin(θ) ,

Let's plug these values in the expression

sec(θ) sin(θ) cot(θ)

= ( 1/cos(θ) ) * ( sin(θ) ) * ( cos(θ) / sin(θ) )

= ( sin(θ) /cos(θ) ) * ( cos(θ) /sin(θ) )

sin cancels out with sin and cos cancels out with cos

So , answer comes out to be

=( sin(θ) /cos(θ) ) * ( cos(θ) /sin(θ) )

= 1

4 0

3 years ago

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Find the equation of the line that is perpendicular to the given line and passes through the given point.

Solnce55 [7]

Answer:

y = 3x + 2

y = 6x – 3

y = 2/3x + 6

y = –1/3x – 4

y = 2x + 1

Correct answer:

y = 3x + 2

Explanation:

Convert the equation to slope intercept form to get y = –1/3x + 2. The old slope is –1/3 and the new slope is 3. Perpendicular slopes must be opposite reciprocals of each other: m1 * m2 = –1

With the new slope, use the slope intercept form and the point to calculate the intercept: y = mx + b or 5 = 3(1) + b, so b = 2

So y = 3x + 2

7 0

2 years ago

2x-8y=-24 6x+4y=68 Using elimination method

nikklg [1K]

6(2x-8y=-24)
-2 (6x+4y=68)
12x-48y=-144
-12x-8y=-136
-56y=-280

-280÷-56=5
y=5

2x-8 (5)=-24
2x-40=-24
+40. +40
2x=16
16÷2=8
×=8

y=5
x=8
hope this helps, if needed a picture for better understanding,please comment.

4 0

2 years ago

If b=4 and c=2, what is 2b-3c?

Paraphin [41]

First, fill in the variables with the numbers given so that
2b-3c
will then become
2(4)-3(2)
next, you refer to the order of operations to simplify the problem down, the multiplication being the first doable operations in this problem.
8-6
and next operation being subtraction.
2
is the final answer

6 0

2 years ago

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If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago