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Is there a =0 at the end?

What you are describing is factoring.

x^2 + 9x

x(x+9) is the answer

11/2 is the answer. You dont even know what kind of values i ended up trying to convert for the formula. All because of a little error. thanks guys

is the answer 11?

Yes, Like gene3141 I get:

Sin(B) =1/sqrt(5), which is the same as: Sqrt(5) / 5

Area of a circle:  A  =  pi · r²     --->     A  =  pi · 7²     --->     A  =  49pi      (approximately 153.94)

I get a different value for sin(B).

If  cos(B)  =  ( 2sqrt(5) ) / 5, I draw the reference right triangle to have an adjacent side (x-side) of 2sqrt(5) and hypotenuse of 5.

Then, using the Pythagorean Theorem, I get the opposite side (y-side) to have a value of sqrt(5),

giving a value for sin(B) to be sqrt(5) / 5.

Yes, thanks so much! 

1 - I'm no expert in Trig.

But, Pi =180 degrees, and 30/180 =1/6Pi, then:

96(Pi)m^2 /6 =16(Pi)m^2

2 - Do the same thing with 2nd question:

20/180 =1/9. and 1/9 =21.5(Pi)Yd^2, so the area of the circle is:

9 x 21.5 x 3.14 =607.59 Yd^2. Makes sense?.

You are quite right.  A typo on my part!

.

4(6x-1)=25x

24x - 4 = 25x

25x - 24x = -4

x = - 4

Using cos² = 1 - sin² 

cos²A = 1 - 5/25 = 20/25   so   cosA = 2√5/5

cos²B = 1 - 1/25 = 24/25   so   cosB = 2√6/5

cos(B-A) = cosBcosA + sinBsinA

sin(B-A) = sinBcosA - sinAcosB

Can you take it from here?

Or this way:

what is a 4A=7A-18

7A - 4A = 18

3A = 18

A = 18/3

A = 6

Did you try the coculator??

Sqrt(169) = 13

Thanks in advance if someone is there.

1/{1.602 x 10^-19} =6.242 x 10^18 electrons = 1 amp/second.

17 minutes x 60 seconds =1,020 seconds.

1,020 x 6.242 x 10^18 =6.367 x 10^21 elecrons that pass a given point in 17 minutes.

Guest#1 beat me to it, but here's my answer anyway:

Verify the following identity:
3/8 - (cos(2 x))/(2) + (cos(4 x))/(8) = sin(x)^4

Put 3/8 - 1/2 cos(2 x) + 1/8 cos(4 x) over the common denominator 8: 3/8 - 1/2 cos(2 x) + 1/8 cos(4 x) = (3 - 4 cos(2 x) + cos(4 x))/8:
(3 - 4 cos(2 x) + cos(4 x))/8 = ^?sin(x)^4

Multiply both sides by 8:
3 - 4 cos(2 x) + cos(4 x) = ^?8 sin(x)^4

cos(2 x) = 1 - 2 sin(x)^2:
3 - 41 - 2 sin(x)^2 + cos(4 x) = ^?8 sin(x)^4

-4 (1 - 2 sin(x)^2) = 8 sin(x)^2 - 4:
3 + 8 sin(x)^2 - 4 + cos(4 x) = ^?8 sin(x)^4

cos(4 x) = 1 - 2 sin(2 x)^2:
3 - 4 + 8 sin(x)^2 + 1 - 2 sin(2 x)^2 = ^?8 sin(x)^4

sin(2 x) = 2 sin(x) cos(x):
3 - 4 + 8 sin(x)^2 + 1 - 2 2 cos(x) sin(x)^2 = ^?8 sin(x)^4

Multiply each exponent in 2 sin(x) cos(x) by 2:
3 - 4 + 8 sin(x)^2 + 1 - 24 cos(x)^2 sin(x)^2 = ^?8 sin(x)^4

cos(x)^2 = 1 - sin(x)^2:
3 - 4 + 8 sin(x)^2 + 1 - 2×4 1 - sin(x)^2 sin(x)^2 = ^?8 sin(x)^4

4 (1 - sin(x)^2) sin(x)^2 = 4 sin(x)^2 - 4 sin(x)^4:
3 - 4 + 8 sin(x)^2 + 1 - 24 sin(x)^2 - 4 sin(x)^4 = ^?8 sin(x)^4

3 - 4 + 8 sin(x)^2 + 1 - 2 (4 sin(x)^2 - 4 sin(x)^4) = 8 sin(x)^4:
8 sin(x)^4 = ^?8 sin(x)^4

The left hand side and right hand side are identical:
Answer: |(identity has been verified)

777! = 1.7974887924634788591539025486099 x 10^1910

actually i was going to go, but then i saw your question so thought to answer it ,

now , \(y=x^2 -2x - 3 \)

now y is also eqal to 1 so 

\(y=1\)

if we quate both of them then,

\(1=x^2 -2x -3 \)

Take 1 to the other side 

\(x^2 -2x-3-1=0\)

\(x^2 - 2x- 4 = 0\)

now it has turned into a simple quadratic equation that you can solve either by factirisation or by the quadratic formula!

ofocourse CPhill, ive been working really hard you know and afterall, im in 10th grade now...so....

x sqrt 3 =6  ?

Divide both sides by sqrt 3

x=  6  /  sqrt 3

x= 6 sqrt 3 /3

x= 2 sqrt 3

x = 3.464

Recall that    e= mc^2

e= 4.36 x 10^?   x  (3x10^8 m/s)^2     Sorry....your question got garbled in posting....I do not know what your numbers are!