Which of the following is equivalent to the expression below?<br> 4x + 3(2x - 1)

Help pls and quick thanks

storchak [24]

Answer:

I think the Question asks to to take w in left side. so, the answer is below

Step-by-step explanation:

w=sa

5 0

2 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

3 years ago

Please help me thank you sooooo much ( also if you take my points <br>I'm reporting you)

Amanda [17]

Answer: x=-10.2 but for me the real answer is x=10.2 but go for x=-10.2

Step-by-step explanation: hope this help

7 0

2 years ago

Read 2 more answers

The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by K(x) 5x/x^2

Olegator [25]

Given :

The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by $K(x) = \dfrac{5x}{x^2+9}$ .

To Find :

Find the time at which the concentration is a maximum. b. Find the maximum concentration.

Solution :

For maximum value of x, K'(x) = 0.

$K'(x) = \dfrac{5(x^2+9)- 5x(2x)}{(x^2+9)^2}=0\\\\5x^2+45-10x^2=0\\\\5x^2 = 45\\\\x = \pm 3$

Since, time cannot be negative, so ignoring x = -3 .

Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6

Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.

Hence, this is the required solution.

5 0

2 years ago

Convert 23° 12' to decimal form

Lina20 [59]

Answer:

15813.64 rounded to nearest hundredth

3 0

3 years ago

Which of the following is equivalent to the expression below? 4x + 3(2x - 1) - 4x