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3 years ago

Solve the following equation. justify each step 0.2x+3.1-2.1x=0.3(x-5)+0.2

1 answer:

4 0

1.) Distribution 0.3(x-5) = 0.3x-1.5

2.) Combine like terms 0.2x and -2.1x ; -1.5 and 0.2

-1.9x + 3.1 = 0.3x -1.3

3.) Add 1.3 to both sides

-1.9x + 4.4 = 0.3x

4.) Add 1.9x to both sides

4.4 = 2.2x

5.) Divide both sides by 2.2 to isolate the variable.

x=2

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There is an ample supply of identical blocks (as shown). Each block is constructed from four 1 × 1 × 1 unit-cubes glued whole-fa

andreev551 [17]

Answer:

336

Step-by-step explanation:

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2 years ago

Find the value of this expression if x = -1 and<br> y=7.<br> Xy / 8

ioda

-7/8

Step-by-step explanation:

$\frac{xy}{8} \\ = \frac{( - 1) \times 7}{8} \\ = - \frac{7}{8}$

So, the value of xy/8 is -7/8

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2 years ago

Which of the following is an example of the commutative property?

anyanavicka [17]

Answer:

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Step-by-step explanation:

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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:

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