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Put all the three boys as one block. In the block, they can be arranged in 3x2x1=6 ways. Then the arrangment for four people (the three girls and one block of boys) is 4x3x2x1=24. 

Multiply 6x24 to get our answer, 144. 

In rectangle \(WXYZ\),
\(A\) is on side \(WX\) such that \(AX = 4\),
\(B\) is on side \(YZ\) such that \(BY = 18\), and
\(C\) is on side \(XY\) such that angle \(ACB= 90^\circ\) and
\(CY = 2CX\).
Find \(AB\).

\(\begin{array}{|lrcll|} \hline &AC^2 &=& 4^2+cx^2 \\ &CB^2 &=& (2cx)^2+18^2 \\ &x^2= AC^2+CB^2 &=& 4^2+cx^2+(2cx)^2+18^2 \\ &x^2 &=& 16+cx^2+ 4cx^2+ 324 \\ &x^2 &=& 5cx^2+ 340 \qquad (1) \\ \hline &x^2 &=& (18-4)^2 + (3cx)^2 \\ &x^2 &=& 14^2 + 9cx^2\\ &x^2 &=& 9cx^2 + 196 \qquad (2) \\ \hline (2)=(1): & 9cx^2 + 196 &=& 5cx^2+ 340 \\ & 9cx^2 - 5cx^2 &=& 340 - 196 \\ & 4cx^2 &=& 144 \\ & cx^2 &=& 36 \\ & \mathbf{cx} &=& \mathbf{6} \\ \hline (1): &x^2 &=& 5cx^2+ 340 \quad | \quad cx=6 \\ &x^2 &=& 5*6^2+ 340 \\ &x^2 &=& 520 \\ &x &=& 2\sqrt{130} \\ &\mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}\)

AB is \(\approx \mathbf{22.8}\)

N=12; R=0.054/12;PV=9000; FV=PV*(1 + R)^N; print"FV =$",FV

FV =$ 9,498.21 - Monthly compound for 1 year, or 12 months.

N=2; R=0.054/2;PV=9000; FV=PV*(1 + R)^N; print"FV =$",FV

FV =$ 9,492.56 - Semi-annual compound for 1 year.

FInd all k for which x^2 + 5kx + k^2 + 5 is divisible by x + 2, but not by x + 3.

Area = ( AP + PS/2 )²   

As follows:

Correction:  The final area should just be 9, of course, not 9/2.

A + 20cm² = 32cm² + 16cm²

A = 28cm²       

It is nice to see that you have finished for yourself ConfusedKitten.  Congratulations!

The first one is quite simple.

Treat it like any other algebraic fraction ONLY always keep the two sides seperate.

Get a common denominator for the RHS then equate cooefficients.

v - p = (2,-7,1).

(a) There are 3^n ways of distributing the elements among A, B, and C, and there are 2^n ways of choosing each subset, so the probability is 3^n/(2^n*2^n*2^n) = 3^n/8^n.

(b) There are 2^n ways of choosing C, then 2^(n - 1) ways of choosing B, then 2^(n - 2) ways of choosing A, so the probability is 2^n*2^(n - 1)*2^(n - 2)/(2^n*2^n*2^n) = 1/8.

AB works out to 4*sqrt(11).

Looking for a pattern, we find that the numbers are decreasing in two ways.

First, starting from the very right (the number 9) to the left.

Second, decreasing from left to right starting from the number 5. 

So we can see through this pattern that the next term is 9876541234568.

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If a quadratic has nonreal roots, then the discriminant \(b^2-4ac<0\).

Plug in the values for b, a, and c into this equation to get \(k^2-4*12*27<0\). 

Rearrange to get \(k^2<4*12*27\). 

Now factorize 4*12*27 and solve to get \(-36 < k < 36\)

Carly, who is 5 inches taller than her brother Bailey, casts a shadow that is a foot longer than his. If Carly is 50 inches tall, what is the length in inches of Bailey's shadow?

Hello Guest!

\(50:(S_B+1)=(50-5):S_B\\ 50S_B=(S_B+1)\cdot (50-5)\\ 50S_B=45S_B+45\\ 5S_B=45\)

\(S_B=9\)

The length of Bailey's shadow is 9 inches.

  !
 

We let x be the cost of a brownie and y the cost of a cookie.

Alvin pays 4x+3y. Milly pays 2x+16y.

Since Milly pays twice as much as Alvin we form an equation.

2(4x+3y)=2x+16y

8x+6y=2x+16y

6x=10y

x=5/3y

So a brownie is 5/3 times as expensive as a cookie.

You're welcome.

x^2-y^2 is the difference of squares: (x+y)(x-y)

So (x+y)(x-y)=63

We already know x-y=3.

3(x+y)=63

x+y=21

We can form a system of equations from this.

x-y=3

x+y=21

We solve and find that x=12 and y=9.

So x^2+y^2=12^2+9^2=144+81=225.

Thank you so much for helping with a lot of my problems!!!

First, we note that r must be positive, since otherwise \(\lfloor r \rfloor + r\) is nonpositive. Next, because \(\lfloor r \rfloor + r\) is an integer and \( \lfloor r \rfloor + r=12.2\), the decimal part of r must be 0.5. Therefore, r=n+0.5 for some integer n, so that \(\lfloor r \rfloor + r\) and \(\lfloor r \rfloor + r = 2n+0.5 =16.5\). Therefore, n=8, and the only value of r that satisfies the equation is \(\boxed{r=8.5}\).

Oh never mind I see your messages. I responded to them.

1. The discriminant of a quadratic is b^2-4ac.

In the equation 5x^2-2x+8, a=5, b= -2, c=8. So plugging in the values of the variables, b^2-4ac=(-2)^2-4(5)(8)=4-160= -156.

2.

 \(\frac{x+4}{x+5}=\frac{x-5}{2x}\\ 2x^2+8x=x^2-25\\ x^2+8x+25=0\)

Now we just have a quadratic equation and we solve to get  \(x=-4+3i, -4-3i\).

You're welcome.