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She  took  the diameter.....she  should have  used  the radius  =  20  /  2    = 10  in

So

pi  ( 10)^2   =   

100 pi   (in)^2         Also  it needs to be (in)^2   instead of just  inches

(3x - 5)  / (6 - 2x)  =  lx - 7l / 5                  

This one may  be  best solved  with a graph....see here : https://www.desmos.com/calculator/73s4myp99b

The solution  is  x ≈  2.188

The obtuse angle is 108 degrees.

Just like what Chris said, PQR is an inscribed angle, and by definition, an inscribed angle is equal to 1/2 the measure of the arc it contains. 

Therefore, 1/2 * 154 is 77, just like NaturalDisaster and CPhill had said. 

Let  the  longer  leg   =  x

The shorter leg  =  x - 8

The hypotenuse  =  x  + 6

By Pythagoras

(x - 8)^2  + x^2  = ( x + 6)^2      simplify

x^2 - 16x + 64  + x^2  = x^2  + 12x +  36

2x^2  - 16x  +  64  =  x^2  + 12x +  36

x^2 - 28x + 28 =  0

Using the  Quad Formla 

x =   ( 28  ±  sqrt  [  28^2  -  4*28]  )  /  2

x ≈  1.04  reject

x ≈   27   =  the longer leg

x - 8  ≈  19 = the shorter leg

x + 6  ≈ 33  the hypotenuse

Correct,  ND !!!!!

Thanks! It was correct! I had the concept of the conjugate from complex numbers, but I didn't know much about the variables itself!

l  ( 5 - 2x)  / ( x + 2)  l   <   1

Set this up as

-1  <  (5-2x) / ( x + 2)  <  1

We  have  two  inequalities

-1  <  ( 5 - 2x)  / (x + 2)                   also             (5-2x)  / ( x + 2)   <  1

-1 ( x + 2)  < 5 - 2x                                              5 - 2x  <   1 (x + 2)                             

-1x  -2  < 5 - 2x                                                  5  - 2x  <  x + 2

x < 5 + 2                                                             5 - 2  <   x + 2x

x <  7                                                                    3  <  3x

                                                                            3/3  <  x

                                                                              1  <  x     ⇒      x  > 1

The solution interval is

1 < x <  7

Hello :))

I'm not quite sure how to do it either, but I'll try my best to help. 

A conjugate is like an inverse, a+ bi conjugate is a - bi. 

The absolute value of a complex number is sqrt(a^2 + b^2), a + bi. 

So we have a complex number (a+bi) where sqrt(a^2 + b^2) = 1. 

z = a + bi

sqrt(a^2 + b^2) = 1

a^2 + b^2 = 1

We're looking for a - bi in terms of z. 

(a-bi)(a+bi) = a^2 + b^2

But we know a^2 + b^2 = 1. 

(a-bi)(a+bi) = 1

We also know that a + bi = z. 

(a-bi)(z) = 1

a-bi = 1/z

Our answer is 1/z hopefully

I hope this helped. :))

=^._.^=

1/2 * 154 = 77 degrees

I hope I am right 

Notice  that f(x)   intersects  the  line  y  = 1   six times

So   c   =1

PRQ is  an inscribed  angle in a circle......its  measure  = 1/2 of  the measure of its  intercepted arc

So  x =  (1/2) (154)  =  77°

Turns out that the Microsoft scientific calculator is a powerful tool. You can calculate 2^2012, then use mod 10 on it to get the answer of: 

nope not going to tell you Try it!

Attn:  itachiuchihakujo

Behave yourself or you will be banned from the forum.

2,4,8,6,2

So every 5th one ends in 2

so 2^2010 ends in 2

so 2^2012 must end in .. and you can finish it.

It is the element that is 6 down and 1 across.  That should be enough info.

They are vertically opposite angles.

If you do not know what then means, google it.

What is the area of the trapezoid shown?

Hello Guest!

\(A=\dfrac{1}{2}(2+2a+2)\cdot h\\ a=2cos60°\\ h=2sin60°\)

\(A=\dfrac{1}{2}(2+2\cdot 2cos60°+2)\cdot 2sin60°\\ \color{blue}A=5.196\)

  !

Where is the diagram?

The United States Postal Service charges an extra $0.11 in postage if the length of an envelope, in inches, divided by its height, in inches, is less than 1.3 or greater than 2.5 For how many of these four envelopes must the extra $0.11 in postage be paid?

What 4 envelopes?   

As I said, you still haven't proofread your question.

What are the solutions to the trigonometric equation on the interval [0,2π)?

2cos^2 x=cos x

Select all correct solutions.

Hello Guest!

\(2cos^2 x=cos x\\ 2cos^2 x-cos x=0\\ cos x=0\\ \color{blue}x\in \{\frac{\pi}2{},\frac{3\pi}{2}\}\)

\(2cosx-1=0\\ cosx=0.5\\ \color{blue}x\in \{\frac{\pi}{3},\frac{5\pi}{3}\}\)

Select all correct solutions.

0

π/6

π/3

π/2

2π/3

5π/6

π

7π/6

4π/3

3π/2

5π/3

11π/6

2π/3

5π/6

π

7π/6

4π/3

3π/2

5π/3

11π/6

  !

thats not helpful

Hmm, this is hard

i believe it is 5