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What is the sum of all possible values of $x$ such that $2x(x-10)=-50$?  

First, multiply it out.                2x² – 20x  =  –50  

Add 50 to both sides              2x² – 20x + 50  =  0  

We don't have to, but let's 

divide both sides by 2               x² – 10x + 25  =  0   

Factor the quadratic                 (x – 5)(x – 5)  =  0  

The only value is                        x  =  5  

Or if your class counts both roots, even though they're the same, then it's 10.  

_.

G(1) is equal to -3 because when x is 1, y is -3.
Now, find G(-3). So go to the x-coordinate of -3, and see what the y-coordinate is. In this case, the y-coordinate is 5.

Now you are left with G(G(G(5))).

Now here, when the x-coordinate is 5, the y-coordinate is also 5. Since you must take the y-coordinate from G(x) = y, you are left with G(G(5))

I believe you know what this means.

first of all, i passed intermediate algebra with a b plus. i also must keep responses short since i broke my other arm and can t press shift. no disrespect intended.

you can also guess and check

Every 20 minutes, it goes down by 1 degree celsius.

That means from 1:40 to 2 it was -7 celsius. 1:20 to 1:40 -6, 1 to 1:20 was -5.

-2 was 12:00 to 12:20

1 was 11:00 to 11:20

4 was 10:00 to 10:20

7 was 9:00 to 9:20 => 8 was 8:40 to 9:00

you can generalize this, but:

-9*C was from 8:20 to 8:40 AM

If you know (x + 7)^2 = 36, then square rooting both sides would get rid of the squared symbol on the left side of the equation. Hence, you can get that x + 7 = +/- 6

You have two cases, x + 7 = 6 and x + 7 = -6.

Solving for both, you obtain: x = -1 and x = -13

75 + 72 + 69 + 66 + ... + 3 + 0 + -3 ... + -78 + -81

If you notice, every single term on the left and right side cancel out except for -78 and -81

-78 - 81 = -(78 + 81) = -159

\(\sqrt{28 + 16\sqrt{3}}\)

If you want to turn that to \(a+ b\sqrt{c}\), here's how.

First the c is obviously a 3, because what else would it be duhhh. 

\(28 + 16\sqrt{3} = (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2\)

\(2ab\sqrt{3} = 16\sqrt{3} \) 

\(ab = 8\)

\(a^2 + 3b^2 = 28\)

If a and b are integers, you can guess and check a little bit. After some testing, you will find that:

\((a, b) = (4, 2)\)

So: \(\sqrt{28+16\sqrt{3}} = 4 + 2\sqrt{3}\)

Sorry, but 3p-+3p-+p = 45 = -p not 7p - 12.

Thus we obtain p = -45. 3p - (+3p) is distributing -1 to 3p, so it become -3p w/o the parenthesis. It turns out that -+ the minus takes priority.

-9x + 15 = 3(2-x)

Distribute => -9x + 15 = 6 - 3x

Add 9x to both sides => 15 = 6 + (-3 + 9)x

Subtract 6 and distribute => 9 = 6x 

x = 3/2

If you subtract 15 from both sides, yes you get -9 because 6 - 15 = -9, but if you subtract 6 from both sides, you get positive 9 on the other side. Combining like terms can give negative numbers.

It is the distance the number is from 0

|x| is used to denote the absolute value of x, if x = -3, then |-3| = 3, because it is 3 units away from 0. |3| = 3. |x| is always positive.

appreciate it

-16t2+40t=0

i solved like his

-16t2=-40

-16/-40 = 2.5t  

why does he second power disappear  

It looks wrong because you left the "t" off the 40t when you took it to the other side.  

it should have looked like this   –16t² = –40t  

Except I'd have first multiplied both sides by –1

                                                        16t² = 40t  

divide both sides by 8t and obtain     2t   =  5  

divide both sides by 2 and obtain        t   =  5/2  =  2.5  

_.

Hi Guest!

\(y=x-a \\ \text{Add a to both sides:} \\ y+a=x-a+a \\ \therefore y+a=x \\ \iff x=y+a\)

I hope this helps!

Square both sides and tidy up.

sin(theta) = 11/25.

x= -1

I let the central crossection of the sphere be  \(x^2+y^2=16^2\)

So the crossection of the top half is    \(y=\sqrt{256-x^2}\)

The table is at y=-16

So the cut is at y=8

Using Pythagoras you will find that the positive end of the cut is at  \(x=8\sqrt3\)

-------

The volume of the sphere section from x=0 to 8sqrt3  will be

\(V=\displaystyle \pi\int_0^{8\sqrt3}\;256-x^2\;dx\\ V=\pi \left[256x-\frac{x^3}{3}\right]_0^{8\sqrt3}\\ V=\pi \left[256*8\sqrt3 -\frac{8^3*3*\sqrt3}{3}\right]\\ V=\pi \left[4*8^3\sqrt3 - 8^3\sqrt3\right]\\ V=\pi \left[3*8^3\sqrt3 \right]\\ \)

The volume of the rectangular p[rism section from  x=0 to x=8sqrt3 will be

\(V=\pi \displaystyle \int_o^{8\sqrt3} 8^2dx\\ V=\pi \displaystyle \left[64x\right ]_o^{8\sqrt3} \\ V=64*8\sqrt3\;\pi\\ V=8^3\sqrt3\;\pi\)

Subtract

Difference in volume is   

 \(V=2*8^3\sqrt3\;\pi\\ V=1024\sqrt3\;\pi\)

This is top and bottom so you can divide by 2 but it is only half of the needed region so you have to multiply by 2

So that is the  volume of the top bit.

You should be able to do at least a bit more by yourself.     

Hello Guest!

\(0.816496580927726\\ =\frac{816\ 496\ 580\ 927\ 726}{1\ 000\ 000\ 000\ 000\ 000} \color{blue}=\frac{4\ 082\ 482\ 900\ 463\ 863}{500\ 000\ 000\ 000\ 000}\ is\ a\ rational\ fraction.\)

  !

If you take its "continued fraction", you should get the following:

(0; 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 1, 2, 1, 2, 14, 2, 6, 2, 1, 8, 5, 1, 1, 1, 2, 1)

Converting the above to proper fractions, you get the following:

1
1 / 3
3 / 4
4 / 7
7 / 11
11 / 62
62 / 507
507 / 569
569 / 1645
1645 / 10439
10439 / 22523
22523 / 325761
325761 / 674045
674045 / 999806
999806 / 2673657
2673657 / 3673463
3673463 / 10020583
10020583 / 43755795
43755795 / 97532173
97532173 / 433884487
433884487 / 965301147
965301147 / 4295089075
4295089075 / 9555479297
9555479297 / 42517006263
42517006263 / 94589491823
94589491823 / 420874973555
420874973555 / 936339438933
936339438933 / 4166232729287
4166232729287 / 9268804897507
9268804897507 / 41241452319315
41241452319315 / 91751709536137
91751709536137 / 408248290463863

408248290463863 / 500000000000000

From the above, it appears to converge to a "rational fraction".

Note: The above was calculated by a computer programm in C++. If you don't understand "continued fractions", then you should consult your Math Teacher.
 

You have not simplified your question, you have just written it properly.  As you should have done int the first place.

Latex coding is greast if you render it properly.  Otherwise it is just coding garbage.

I know that if x and the denominator are relatively prime then

x/9 will have  one number repeating

x/99 will have 2

x/999 will have 3 digits repeating

..

x/99999999  will have 8 digits repeating

Now

factor(99999999) = 3^2 *11 * 73 * 101 *137

So n is going to have to be a multiple of 1 or more of these factors

1/3 only has 1 repeating digit

1/9 only has 2 repeating digitsw

1/11 only has 2 repeating digits

1/33 only has 2 repeating digits

1/99 only has 2 repeating digits

1/73 = 0.01369863 with all 8 digits repeating.

So the smallest n is 73