Help please <br> I need help asap

Ivanshal [37]

Y + 4x = 8
y = 8 - 4x
substitute 8 - 4x for y in the other equation.
5x + 2(8 - 4x) = 13
5x + 16 - 8x = 13
-3x = -3
x = 1
y = 8 - 4(1) = 4

4 0

3 years ago

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

2 years ago

Find the P-value for the hypothesis test with a standardized test statistic z. Decide whether toreject the null hypothesis for t

iren2701 [21]

Answer:

1.) We don't reject the null

2.) We reject the Null

3.) We do not reject the null

Step-by-step explanation:

Obtaining p values using test statistic:

Given that ; we have a standardized test statistic and α - values ;

Let's define the decision region :

If Pvalue < α ; Reject H0 ; otherwise, fail to reject H0

A.)

Left tail , Z = - 1.32 ; α = 0.10

We can use the Pvalue calculator from Z score :

Pvalue = 0.934

Pvalue > α ; Hence, we fail to reject the null, H0

B.)

Right-tailed test, z = 2.46, α = 0.01

Pvalue from Zscore calculator ;

Pvalue = 0.0069

Pvalue < α ; Hence, we reject the null, H0

C.)

Two-tailed test, z = -1.68, α = 0.05

Pvalue from Zscore calculator ;

Pvalue = 0.093

Pvalue > α ; Hence, we fail to reject the null, H0

6 0

2 years ago

Suppose you choose a marble from a bag containing 5 red marbles, 3 white marbles, and 6 blue marbles. You return the first marbl

Aneli [31]

$|\Omega|=14^2=196\\
|A|=5\cdot6\cdot2=60\\\\
P(A)=\dfrac{60}{196}=\dfrac{15}{49}$

5 0

3 years ago

A coupon subtracts $17.95 from the price p of a pair of headphones. You pay $71.80 for the headphones after using the coupon. Wr

Volgvan

$264.75 I like math it took 15 minutes but I got it

7 0

3 years ago

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Help please I need help asap