All Answers 358122 Answers

The sum of all possible vaues of z is 2 + 3 + 4 + 6 = 15.

So that makes p(x) = x^2 + 2x + 1.

28 is the answer

But this is incorrect...

\(\boxed {800}\)

thank you so much!!

\(\frac{1}{6}\cdot\frac{7}{8}=\frac{7}{48}\)

1/6 for the probability of  rolling a 6, 7/8 is the probability of not rolling a 6.

\(\frac{1}{8}\cdot\frac{5}{6}=\frac{5}{48}\)

1/8 for the probability of  rolling a 6, 5/6 is the probability of not rolling a 6.

\(\frac{7}{48}+\frac{5}{48}=\frac{12}{48}=\frac{1}{4}\)

Add the probabilities together.

There are 1680 possible combinations for pulling out marbles. (8*7*6*5)

There are 6 possible combinations for getting 2 red and 2 blue. (RRBB, RBRB, BRBR, BBRR, BRRB, and RBBR)

\(6\over1680\)\(\Rightarrow\)\(1\over280\)

The answer is \(1\over280\)

If I understand your question:

How many integers are there between 10 and 99?

99 - 10 + 1 = 90 integers

How many 2-digit integers are there that are the same? Well, you have: 11, 22, 33, 44, 55, 66, 77, 88, 99 =9 integers

So, Probability of picking an integer with the same 2 digits =9 / 90 = 1 / 10.

5C4 + 5C5 =5 + 1 = 6 / 2^5 =3 / 16 =0.1875 probability of at least 4 tails

Not math but ok....

365 days..

I can find  cos(A/C) 

Let v be speed of current.

Let t be time to travel 4 miles downstream.

time = distance/speed

t = 4/(10 + v).    (1)

t + (10/60) = 4/(10 - v).    (2)

You now have two equations in two unknowns.  Solve them to find v.

A box has a total surface area of 100. The length of the box is equal to twice its width, as well as equal to 3 less than its height. What is the height of the box?

x * 2x * ( 2x + 3 ) = 100           ( x  is the width)

x = 2.5

The height is:      2x + 3 = 2 * 2.5 + 3 = 8

On \(\triangle ABC\), BC is extended to E such that \(CE=\dfrac{BC}{2}\).

IF D is the mid point of AC and ED cuts AB at F,

Find the value of \(\dfrac{AB}{AF}\).

Ceva's theorem

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{BC}{CE}*\dfrac{EG}{GA}*\dfrac{AF}{BF}} &=& \mathbf{1} \\\\ \dfrac{2}{1}*\dfrac{EG}{GA}*\dfrac{AF}{BF} &=& 1 \\\\ \dfrac{EG}{GA} &=& \dfrac{1}{2}* \dfrac{BF}{AF} \\\\ \mathbf{\dfrac{GA}{EG}} &=& \mathbf{2* \dfrac{AF}{BF}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{AD}{DC}} &=& \mathbf{\dfrac{GA}{EG}+\dfrac{AF}{BF}} \\\\ \dfrac{1}{1} &=& 2* \dfrac{AF}{BF}+\dfrac{AF}{BF} \\\\ 1 &=& 3* \dfrac{AF}{BF} \\\\ \dfrac{AF}{BF} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{BF}{AF}} &=& \mathbf{\dfrac{3}{1}} \\ \hline BF &=& \dfrac{3}{4}AB \\ AF &=& \dfrac{1}{4}AB \\\\ \dfrac{AB}{AF} &=& \dfrac{AB}{\dfrac{1}{4}AB } \\\\ \mathbf{\dfrac{AB}{AF}} &=& \mathbf{4} \\ \hline \end{array}\)

A triangle has side lengths 3, 4, and 5.  Find the radius of the inscribed (triangle) circle.

Semiperimeter is:         s = (3+4+5)/2 = 6

Area of a triangle is:     A = 1/2 (3*4) = 6 u²

Inradius r is:                     r = A / s = 1  

x+ (x+2) + (x+4) + (x+6)=68
4x+12=68
4x+12-12=68-12
4x=56
dividing both side by 4,
x=14
what are the numbers: 14,16,18,20. Add them and you'll get 68
BUT, THE QUESTION IS WHAT IS TWICE THE SMALLEST INTEGER so,
Twice the smallest integer is
2x= 2(14)
x=28

 

If 6 Dogs can find 6 bones in 6 minutes, how long will it take 100 Dogs to find 100 bones?

Hello Guest!

\(W_1=N\cdot n_1\cdot t_1\\ N=\frac{W_1}{ n_1\cdot t_1}=\frac{6\ bones}{6\ dogs\ \cdot \ 6\ min}\)

\(N=\frac{1}{6}\cdot\frac{bones}{dog\cdot min}\)

\(t=\frac{W_2}{N\cdot n_2}\\ t=\large \frac{100}{\frac{1}{6}\cdot 100}\cdot \large \frac{bones}{ \frac{bones}{dog\cdot min}\cdot dogs}\)

\(t=6\ min\)

  !

∠P = 60º       ∠Q = 90º       ∠R = 120º        ∠S = 90º

∠P + ∠R = 180º            ∠Q + ∠S = 180º      

Let n and k be positive integers such that n<10^6  and
\(\dbinom{13}{13} + \dbinom{14}{13} + \dbinom{15}{13} + \dots + \dbinom{52}{13} + \dbinom{53}{13} + \dbinom{54}{13} = \dbinom{n}{k}\).

Find the value of n and k.

Solve system  \(\dfrac{1}{x} - \dfrac{1}{y} = 1,\ 2x + y = 7xy \quad| \quad x \ne 0,\ y\ne 0\).

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x} - \dfrac{1}{y}} &=& \mathbf{1} \quad| \quad *xy \\\\ \dfrac{xy}{x} - \dfrac{xy}{y} &=& xy \\\\ y - x &=& xy \\ y &=& xy +x \\ y &=& x(y +1) \\\\ \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{2x+y} &=& \mathbf{7xy} \quad | \quad \mathbf{x=\dfrac{y}{y+1}} \\\\ 2*\dfrac{y}{(y+1)}+y &=& 7*\dfrac{y}{(y+1)}*y \\\\ \dfrac{2y}{(y+1)}+y &=& \dfrac{7y^2}{(y+1)}\quad| \quad *\dfrac{(y+1)}{y} \\\\ 2+y+1 &=& 7y \\ 3 &=& 6y \\\\ y &=& \dfrac{3}{6} \\\\ \mathbf{y} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \quad | \quad y=\dfrac{1}{2} \\\\ x &=& \frac{1}{2}\above 1pt \frac{1}{2}+1 \\\\ x &=& \frac{1}{2}\above 1pt \frac{3}{2} \\\\ x &=& \dfrac{1}{2}*\dfrac{2}{3} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

In triangle ABC, AP = PQ/2 = BQ and CR/RA=3/2.
Find the area of BQC/(the area of)CRQ.

In triangle

\(\text{In triangle } ABC,\ AP = \frac{PQ}{2} = BQ,\quad\dfrac{CR}{RA} = \dfrac{3}{2}. \dfrac{[BQC]}{[CRQ]}. \)