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Here is my attempt :

a)  3^4 = 81 cards

b)  81C3 = 85320 sets of 3 cards

c)    All different       3! *3!*3!  =  216

d)  Say they are all red  and the other attributes are all different  3!*3!*3!   = 216   

But there are 12 different things that could be the same so that is  12*216 = 2592

e)    exactly 2 matching atributes

Say they are all R1  they are each allotted a shape.  then there is 3! choices for shading.

12C2 *3!  = 66*6 = 396

f)   say they are all the same except one is red, one Purple, and one green.  There is one set of those.

So there is one set  for each attribute.  so that is 4 

Let's try to work backwards.

Before Ligma ate 5 nuts, there were 2+5=7 nuts.

So Adi and Sugma ate 20-7=13 nuts.

Using Guess and check,

we can find that Adi ate 8 nuts and Sugma ate 5.

Answer: 8 nuts

Assuming that the middle line cuts the trapezoid into a square and a triangle,

Perimeter=13+13+19+10=55ft

\(6C3 * (\frac{m}{n})^3 *(1-\frac{m}{n})^3=\frac{4^4}{5^5}\\\)

Now solve it

Hint:

Pythagoras's theorum

Have you tried drawing it?

there is no question here                                                            

9*8 = 72  

That is assuming the chilren are not identical and that it is possible for one to get none.

Vieta's formula allows you to calculate the sum or product of the roots of the quadratic equation. 

For example, let's say we have the quadratic \(x^2 + 6x + 8\) (in the form \(ax^2 + bx + c\)). By Vieta's, the sum of the roots is \({-b \over a} = {-6 \over 1} = -6\), and the products of the roots is \({c \over a} = {8 \over 1} = 8\)

Hope this helps!!

\(9+\frac{27}{x}+\frac{8}{x^2} = \frac{7}{x^2} + \frac{24}{x} + 5\)

Multiply both sides by x²,

\(\Rightarrow 9x^2 + 27x + 8 = 7 + 24x + 5x^2\)

\(\Rightarrow 4x^2 + 3x +1 =0\)

\(=\frac{- 3\pm \sqrt{9-16}}{8} = \frac{-3 \pm \sqrt{7}i}{8}\)

\(=\boxed{ \frac{-3 + \sqrt{7}i}{8} , \frac{-3-\sqrt{7}i}{8}}\)

\((2y-1)(4y^{10}+2y^9+2y^8+y^7)\)

\(= \boxed{8y^{11}+2y^9-y^7}\)

Use the formula \(\frac{f(x+h)-f(x)}{h}\) to find the derivative. 

\(\frac{f(x+h)-f(x)}{h}= \frac{[5(x+h)^3-11(x+h)^2+7(x+h)+3]-[5x^3-11x^2+7x+3]}{h}\)

\(=\frac{5x^3+15x^2h+15xh^2+5h^3-11x^2-22xh-11h^2+7x+7h+3-5x^3+11x^2-7x-3}{h}\)

\(=\frac{15x^2h+15xh^2+5h^3-22xh-11h^2+7h}{h}\)

\(= 15x^2+15xh + 5h^2 -22x-11h+7\)

Substitute the \(h\) with 0, \(= \boxed{15x^2-22x+7}\)

oh I get it now, x = a+b+c, and when y = 1, x = 1( according to the points) so since x = a+b+c, a+b+c = 1

they didn't do their latex correctly, it was \(\frac{x}{4} - \frac{y}{5}=1\)

90/0.42 = (90 * 100)/42 = 214.28 ≈ 214

Thanks for the help!

The probability of getting 2 red marbles is \({3 \over 3 + k} \times {2 \over 2 + k} \)

The probability of getting 2 white marbles is \({k \over k + 3} \times { k - 1 \over k + 2}\)

This means that we have the equation \(\left(\frac{3}{3+k} \times \frac{2}{2+k}\right)+\left(\frac{k}{k + 3}\times \frac{k-1}{k+2}\right)=0.5\)

Now, just solve for k...

We have this equation: \(\frac{x}4-\frac{y}5=1\)

Subtract \({x \over 4}\) from both sides: \(-{y \over 5} = 1 - {x \over 4}\)

Multiply the entire equation by -5: \(y = {5 \over 4} x - 5\)

This means that the slope is \({5 \over 4}\), the slope of the line that is perpendicular to this is the negative reciprocal, or \(\color{brown}\boxed{-{4 \over 5}}\)

Ok!

What is the slope of a line perpendicular to the line whose equation is (x/4)-(y/5)=1? Express your answer as a common fraction.

We have the system: 

\(2x + 5y = 11 \ \ \ \ \ \ \ (i)\)

\(3y = x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)\)

Start by substituting \((ii)\) into \((i)\), which gives us \(2(3y) + 5y = 11\), meaning \(y = 1\)

Now, plug in \(y = 1\) into \((ii)\), to get \(3 \times 1 = x\), meaning \(x = 3\).

This means that the solution is \((3,1) \), meaning the sum is \(3 + 1 = \color{brown}\boxed{4}\)

Believe it or not, I don't speak LaTex