All Answers 358479 Answers

That answer is wrong.

I presume the boxes and inequalities look like such 

[ ] < [ ] > [ ] < [ ] > [ ]

We can set the inequality as

a < b > c < d > e

To solve the question, there are 2 cases:

case 1:

b=4 d=5 (or the other way around)

Then we can set a, c, and e, as any number (1, 2 or 3)

so for case 1 we have 2*3!=2*6=12

case 2:

b=5, a=4 (or on the other side, d=5, e=4)

In this case we know that d has to be 3, other wise the inequality won't *work*

we have two ways to put 1, and 2 into the equation, so 

For case 2, there are 2*2=4 ways

Add them together: 4+12=16

(wow this sounded a lot simpler in my head) Please confirm this

Your answer is wrong.

I think this question was already answered here by gb1falcon

 https://web2.0calc.com/questions/need-help-with-system_6 

google translates  分解因式咋弄   as   How to factorize   

google translates  x的三次方+2x   as   x to the third power + 2x  

I assume they want      x³ + 2x  =  (x)(x² + 2)        

N=2*4; R=0.002/4;PV=4000; FV=PV*(1 + R)^N

a -  FV =$ 4016.03 - balance of the account after 2 years

b - $4016.03  -  $4000 =$16.03 - total interest earned  after 2 years.

2  x  6! =1440

2^4 * 3 * 7=336

1440  +  336 =1776 ways of arranging 4 pairs of siblings around the table.

Not bad, but a simple arithmetic mistake did you in.   

Actually, there are two mistakes, both subtraction.  

So, we can conclude:

x=y+10

x+25=2y-50

x=2y-25                              x  =  2y – 75  

y+10=2y-25               y + 10  =  2y – 75  

y=2y-15                             y  =  2y – 85  

y=15                                –y  =  –85   

                                          y  =  85   

So, your brother has $85.  

_.

Thanks, that's right!

First, we count the number of ways to fill the offices if Alex and Bob are both officers. There are 18 ways to choose the President, 17 ways to choose the Vice President, and 16 ways to choose the Treasurer, for a total of 18×17×16=4896.

Next, we count the number of ways to fill the offices if Alex is not an officer. There are 19 ways to choose the President, 18 ways to choose the Vice President, and 17 ways to choose the Treasurer, for a total of 19×18×17=5832.

Finally, we subtract the number of ways to fill the offices if both Alex and Bob are officers from the number of ways to fill the offices if Alex is not an officer, to get 5832−4896=936​.

10562 feet

Let's set variables. You will be x, and your brother will be y.

x-5=y+5

x+25=2(y-25)

So, we can conclude:

x=y+10

x+25=2y-50

x=2y-25

y+10=2y-25

y=2y-15

y=15

So, your brother has $15.

Your answer is wrong.

We can write [(1 + \sqrt{-3})^6 = (1 + i \sqrt{3})^6.]Let z=1+i3​. Then z2=1+2i3​+3=4i3​, so z3=−3i, and \begin{align*} z^6 &= z^3 z^3 \ &= -3i(-3i) \ &= 9. \end{align*}

Therefore, (1+i*sqrt(3))^6 = 9 ​.

Triangle ABC is a right triangle, so by the Pythagorean Theorem, AC=122+42​=13. Then by the Pythagorean Theorem in triangle ABX, BX^2 = 12^2 - 4^2 = 144 - 16 = 128, so BX= sqrt(128) = 8*sqrt(2).

If the season consists of 100 games, we have played 75 games and have 25 games left to play. We have won 45 games, so we have lost 55 games. To finish the season with the same number of wins as losses, we must win 55 games. Thus, we must win 55/25 = 22% of the remaining games.

Therefore, the answer is 22​.

Please post worthwhile questions on this forum

i edited it...

Hi,

No, a,b, and c are just letters for the "coefficients"
I.e.: 
Any quadratic equation can be written as: \(ax^2+bx+c\)

And, in general, the discriminant is \(b^2-4ac\).

So for example (Try to solve these)

\(2x^2-4x+3\)

Here, a=2, b=-4, c=3

\(7x^2-\dfrac{13}{2}x-13=0\)

Here, a=7, b=-13/2, c=-13

Thanks for your help. I do have one question. would a and b = x,y?

????

\(\text{Hello ZBRS7311,} \\ \text{This question is about system of two quadratic equations, and it asks about which values of c does this system have: } \\ \text{1-Exactly one real solution} \\ \text{2-More than one real solution} \\ \text{3- No real solutions} \\ \text{The strategy is to make use of the discriminant of a quadratic equation. Namely the quantity: } \\ \Delta = b^2-4ac \space\space\space \text{Where } \Delta \space\text{is called the discriminant}\\ \text{But the problem is, which quadratic equation shall we apply this on? } \\ \text{Well, the quadratic equation that satisifies both "constraints" I.e. the intersection of both equations, I.e. equate both equations} \\ \text{Hence, setting both equations equal to each other, (eliminating y), yields:} \\ \implies 6x^2-9x+c=5x^2-3x \\ \iff x^2-6x+c=0 \\ \text{Notice: a=1,b=-6,c=c} \\ \text{Now for part (a) of the question, to have one real solution, the discriminant must be equal to zero} \\ \iff (-6)^2-4(1)(c)=0 \implies c=9 \\\text{Thus, this is the value of c such that the system has exactly one pair of (x,y) I.e. one solution}\\ \text{Next, for part (b), we want the system to have multiple solutions, this will be satisifed if and only if the discriminant is greater than 0} \\ \iff (-6)^2-4(1)(c)>0 \implies 36-4c> 0 \implies c<9 \\ \text{Thus, whenever c<9, the system will have more than one real solution} \\ \text{For the last part, (c), for the system to have no solutions, the discriminant must be less than zero} \\ (-6)^2-4(1)(c)<0 \implies 36-4c<0 \implies c>9 \text{Therefore, to have no real solutions, c>9} \)I hope this helps, and do not hesitate to ask further questions.