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zalisa [80]
3 years ago
14

Solve the equation Z-19=34

Mathematics
1 answer:
Ede4ka [16]3 years ago
5 0

Answer:

z=53

Step-by-step explanation:

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Please help? I’m super lost...
babunello [35]

Answer:

Step-by-step explanation:

In all of these problems, the key is to remember that you can undo a trig function by taking the inverse of that function.  Watch and see.

a.  sin2\theta =-\frac{\sqrt{3} }{2}

Take the inverse sin of both sides.  When you do that, you are left with just 2theta on the left.  That's why you do this.

sin^{-1}(sin2\theta)=sin^{-1}(-\frac{\sqrt{3} }{2} )

This simplifies to

2\theta=sin^{-1}(-\frac{\sqrt{3} }{2} )

We look to the unit circle to see which values of theta give us a sin of -square root of 3 over 2.  Those are:

2\theta =\frac{5\pi }{6} and

2\theta=\frac{7\pi }{6}

Divide both sides by 2 in both of those equations to get that values of theta are:

\theta=\frac{5\pi }{12},\frac{7\pi }{12}

b.  tan(7a)=1

Take the inverse tangent of both sides:

tan^{-1}(tan(7a))=tan^{-1}(1)

Taking the inverse tangent of the tangent on the left leaves us with just 7a.  This simplifies to

7a=tan^{-1}(1)

We look to the unit circle to find which values of <em>a</em> give us a tangent of 1.  They are:

7\alpha =\frac{5\pi }{4},7\alpha =\frac{\pi }{4}

Dibide each of those equations by 7 to find that the values of alpha are:

\alpha =\frac{5\pi}{28},\frac{\pi}{28}

c.  cos(3\beta)=\frac{1}{2}

Take the inverse cosine of each side.  The inverse cosine and cosine undo each other, leaving us with just 3beta on the left, just like in the previous problems.  That simplifies to:

3\beta=cos^{-1}(\frac{1}{2})

We look to the unit circle to find the values of beta that give us the cosine of 1/2 and those are:

3\beta =\frac{\pi}{6},3\beta  =\frac{5\pi}{6}

Divide each of those by 3 to find the values of beta are:

\beta =\frac{\pi }{18} ,\frac{5\pi}{18}

d.  sec3\alpha =-2

Let's rewrite this in terms of a trig ratio that we are a bit more familiar with:

\frac{1}{cos(3\alpha) } =\frac{-2}{1}

We are going to simplify this even further by flipping both fraction upside down to make it easier to solve:

cos(3\alpha)=-\frac{1}{2}

Now we will take the inverse cos of each side (same as above):

3\alpha =cos^{-1}(-\frac{1}{2} )

We look to the unit circle one last time to find the values of alpha that give us a cosine of -1/2:

3\alpha =\frac{7\pi}{6},3\alpha  =\frac{11\pi}{6}

Dividing both of those equations by 3 gives us

\alpha =\frac{7\pi}{18},\frac{11\pi}{18}

And we're done!!!

8 0
3 years ago
No links or spam i will report! please help thank you!&lt;3
ArbitrLikvidat [17]

Answer:

A. The two lines are neither parallel nor perpendicular.

Step-by-step explanation:

8 0
3 years ago
What is the solution set of the given system?
AVprozaik [17]

What is the system? (Is there supposed to be an image?)

3 0
3 years ago
What is the inverse of y = x+3/4 ?<br><br> a. y=x-3/4<br> b. y=4x-3<br> c. y=4x+3<br> d. y= x-3/-4
viva [34]
Y = x + 3/4

x = y - 3/4

swap the variables:-

y = x - 3/4  (answer).   Its a.
4 0
3 years ago
Read 2 more answers
What is the remainder when 6x+12 is divided by 2x-8.​
bagirrra123 [75]

Answer:

4(3x-2)

Step-by-step explanation

6x+12/2x-8

6x+6x-8

12x-8

4(3x-2)

7 0
3 years ago
Read 2 more answers
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