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Coin \(A\)can either have more heads or less heads than Coin \(B\), but not both.

The total number of possibilities for the coin flips is \(8\cdot 4=32\).

Subtracting the number of flips that have less heads for coin \(A\)or an equal amount of heads makes \(\frac{16}{32}=\boxed{\frac{1}{2}}\).

the answer is 12

If each person shakes hands with exactly one other person, then there will be 23/2 handshakes, since two people are required to complete a handshake. This is equal to 11.5 handshakes, which is clearly impossible. We can achieve 12 handshakes by forming two rows of 11 and 12 people. Each person in the first row shakes hands a different person in the second row. This will give eleven handshakes. The leftover person must shake hands with someone, giving 12 handshakes total.

If Sin(A) = 2/3, then:

2/3 =BD / 24

BD =16

If sin(C) = 3/4, then:

3/4 = 16 / BC

BC =64 / 3 =21 1/3

V =Pi x r^2 x h/3

Circumference =Radius x 2 x pi

16 pi =Radius x 2 x pi

Radius = 8 inches 

V =Pi x r^2 x h/3

192 pi = pi x 8^2 x h/3, solve for h

h = 9 inches - the shorter height.

The ratio is 9 : 30 = 9/30 = 3/10

Using the formulas for "identical" balls into "distinct" boxes listed here in Formulas 5, 6, 7, 8

Seems to give different results:

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

Angle ABC = 180 - 56 - 53 =71 degrees.

Sin(71) / 11 =Sin(53) / BC

BC = 9.29

The balls are the stars. There are 6 of them and they are all the same.

* * * * * *

Now I will assume  there can be zero balls in one or even two of the containers.

I have the containes in a row. Doesn't matter what order.  Say red, blue green.

The bars represent dividers between the boxes. there will be 2 dividers

the red cont gets the balls on the left, the blue gets the middle balls and the green gets the remaining balls

These are the possible position of the bars (remembering that any 2 of them will be chosen.

||*||*||*||*||*||*||

So there are 14 bars.

How many ways can 2 of them be chosen.

14C2 = 91 ways

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Of course if each container must have at least one ball then the problem is a bit different.

*|*|*|*|*|*

The number of ways would be  5C2 = 10 ways

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I think that is right. 

xy= 6 - 2x - 3y................(1) 

 yz= 6 - 4y - 2z...............(2)                                                                                                                                     
xz= 30 - 4x - 3z..............(3)
From (3) above:
z = (30 - 4 x)/(x + 3)
Sub z into (2) above
y(30 - 4 x)/(x + 3)= 6 - 4y - 2[(30 - 4 x)/(x + 3)]...........(4)
xy= 6 - 2x - 3y..............(1), solve for x, y in (1) and (4)
x = - 9 and y = -4, or:
x = 3  and y = 0
Sub x values into (3) above:
z = 3 and z = -11  

\(\dfrac{x}{3} + 1 = \dfrac{x + 3}{x}\\ \dfrac{x+3}{3} = \dfrac{x + 3}{x}\\ 3x*\dfrac{x+3}{3} = 3x*\dfrac{x + 3}{x}\\ x(x+3)= 3(x+3)\\ x^2+3x=3x+9\\ x^2=9\\ x=\pm 3\)

Nevermind; I managed to solve these.

ok, I see your point,

The word 'simultaneously' is  not only superfluous it is also misleading.

Hey Guest!

\(\text{volume of the cube} - \text{volume of the cylindrical section} = \text{volume of the remaining solid.}  \)

The volume of the cube: \(\text{length} * \text{width} * \text{height} = 4^3 = 64\)

Volume of the cylinder: \(π * \text{radius}^2 * \text{height} = 22π * 4 = 16π\)

\(64-16π=13.73251754\cdots\)

I hope this helped,

Gavin. 

Find all real numbers that satisfy the equation: \(\frac{1}{r^3+7}-7=\frac{-r^3}{r^3+7}\).

\(\frac{1}{r^3+7}-7=\frac{-r^3}{r^3+7}\\ \frac{1}{r^3+7}-\frac{-r^3}{r^3+7}=7\\ \frac{1+r^3}{r^3+7}=7\\ 1+r^3=7r^3+49\\ 6r^3=-48\\ r^3=-8\\ r=\boxed{-2} \)

I hope this helped,

Gavin

(x+4)^2 makes x^2+8x+16 so it will be (x+4)^2-16-1=0 which means (x+4)^2=17. b=17

A binomial is (a+b)^2

when you multiply it out you have

a^2  + 2ab + b^2 

we can plug in y=a

24 / 2 = b = 12

12^2 = 144

144 - 16 = 128

thus k = 128

Like if it helped  Thanks!

In the line 4x+ 7y + c = 0, the sum of the a- and y- intercepts is 22 . Find c.

Points $A, B, C$ and $D$ have these coordinates: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$ and $D(-3,0)$. What is the area of quadrilateral $ABCD$?

Find the value of b in -x^²+bx-5<0 where x is element (-infinity,1) (1, infinity)?

x² + 25x + c   c = ?

25x : 2x = 12.5

12,5² = 156.25

c = 156,25

(x² + 12,5)²