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Vertex at 0,0     and includes  20,20   obviously is opening UPWARD....like a bowl

-b/2a  = 0     so b = 0                h, k = 0 , 0

ax^2    = 20

a20^2  = 20

a = 20/400 = .05

y = .05 x^2                  or     y= .05(x-0)^2   + 0

Graph:

https://www.desmos.com/calculator/f5ghmltx1e

Nice job! I can confirm because I got the same answer!

srry for the mess, this is my first post

XD

3.  Divide   f(x)  into the product   f *g    to find  g(x) 

4.   I'm not sure I understand this question      f*g  =  g*f       ??

                -3x (x^2/(2x-3))   =     -3x^3 / (2x-3)      x cannot equal 1  1/2     otherwise all  x   ??

Here's the answer: 10

6xyz+30xy+21xz+2yz+105x+10y+7z=812

We can group them and factor:

6xy(z+5)+21x(z+5)+2y(z+5)+7z=812

(z+5)(6xy+21x+2y)+7z=812

Now if we add 35 to both sides, we can put 7z into parenthesis too,

(z+5)(6xy+21x+2y)+7z+35 --> 7(z+5)=812+35

(z+5)(6xy+21x+2y+7)=847

This simplifies to

(z+5)(3x+1)(2y+7)=847.

Now we can find the factors of 847, which are:

1, 7, 11, 77, 121, 847.

Now, we must find three positive integer values that multiply to 847. There are three of them, which are

and 7x11x11(they don't have to be distinct)

(z+5)(3x+1)(2y+7) take the place of the three integers.

Now we must test all three possibilities.

z+5=1

z=-4

Since it is negative, 1 does not work

3x+1=1

3x=0

x=0

But zero is neither positive or negative, so 1 does not work for this one either

2y+7=1

2y=-6

y=-3

Since this is also negative, it does not work

This means that 

1x7x121

1x11x77

do not work.

That means that (z+5)(3x+1)(2y+7)=7x11x11

BUT...

Which is which?

We must test again...

z+5=7

z=2 This works...

3x+1=11

3x=10

x=10/3

This does NOT work because z must be an integer.

So

z+5 must equal 11

z=6

Now we need to find the other 11 and the 7

This is easy, because we already know that 3x+1 cannot equal 11, so it equals 7.

3x+1=7

3x=6

x=2

This means that

2y+7=11

2y=4

y=2

NOW WE JUST ADD THEM UP

2+2+6=10

The answer is 10

XD

I'm not the best at counting and probability problems, so don't get mad if I get this question wrong.

1. There are \(4*P{5 \choose 2}=240\) ways to place an economics text book in the middle as well as place a biology textbook at each end.

2. There are \(6!\) ways to arrange the remaining text books.

So \(720*240=172800\) total ways.

Hello SpongeBobRules24! I am mentally lazy so I'm just gonna manually solve for a, b, c

Let us subsitute:

P(1) = a + b + c = 1

P(2) = 4a + 2b + c = 2

P(3) = 9a + 3b + c = 4

We have:

(1) a + b + c = 1

(2) 4a + 2b + c = 2

(3) 9a + 3b + c = 4

We multiply (2) by 2, and we keep (3)

(1) a + b + c = 4

(2) 8a + 4b  + 2c = 4

(3) 9a + 3b + c = 4

(2) - (3)

-a + b + c = 0

a = b + c

Substitute that into (1)

2a = 1

a = 1/2

Let's go back to the original equations:

(1) a + b + c = 1

(2) 4a + 2b + c = 2

(3) 9a + 3b + c = 4

Substitute a:

(1) b + c = (1/2)

(2) 2b + c = 0

(3) 3b + c = -(1/2)

We do (3) - (2):

b = -(1/2)

We plug that into (1)

c = 1

a - b - c = (1/2) - -(1/2) - 1

Which evaluates to \(\boxed{0}\)

\(\text{Cylinder volume formula}={\pi}r^2h=V\)

In other words, we are taking a circle and multipying it by the height of the cylinder.

_________________________________________________________________

Since the second cylinder has twice the height and twice the radius, we plug that into the cylinder formula.

(1) \(\pi(2r)^22h={\pi}4r^22h\)

Since \({\pi}r^2h=V\), we divide by the cylinder formula to see how much V has changed.

1. \(\frac{{\pi}4r^22h}{{\pi}r^2h}=\frac{?}{V}\)

Canceling out, we get:

2. \(\frac{8}{1}=\frac{?}{V}\)

\(?=8V\)

The volume of the second cylinder is \(\boxed{8V}\)

Of course, you can always substitute real values, but its more fun for me to solve it algebraically.

a 44%+24%=68%

b 68%+13%=81%

c 100%-81%=19%

Hello SpongeBobRules24!

Distribute:

(1) \(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

Combine like terms (fractions with the same denominator):

(1) \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)

Consider:

It was given that 

(2) \(a+b+c=0\)

We solve for the numerators of the three fractions in (1).

Solving:

Let us do the fraction with \(a\) as the denominator.

\(b+c=-a\)

We substitute this into (1):

\(\frac{-a}{a}=-1\)

Our equation is now:

\(-1+\frac{a+c}{b}+\frac{a+b}{c}\)

We repeat the process of solving for the numerators of the fractions in (1) and eventually get:
\(-1+-1+-1\)

Which evaluates to \(\boxed{-3}\)

That is all part of one question, so I don't think that is all of their homework.

The only game I play that is similar to RoBeats is Geometry Dash.

My username is SpongebobRules26.

I also used to play Phantom Forces, but it kind of got annoying because it was so hard to level up, and there were so many people attacking you at once.

I mostly play Ninja Legends, Jailbreak, and Strucid.

140 is also wrong, but thanks for trying.

Ohh, haha nice! Do you still play?

The total length of the intervals is 168*pi.

See https://web2.0calc.com/questions/i-m-really-stumped-polynomials#r1

The distance between the foci is 3*sqrt(2).

|AP - BP| = 4.

We are not interested in doing your homework for you.

(5,2)

1

1/3+2/3=1