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Oh, my drawing didn't attach. Oops.

Multiples  of 2

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100

So

Multiples of 2  that  are not also multiples of  4 or 5  are

2 6 14  18  22  26   34  38  42   46   54   58   62    66    74  78  82  86   94   98

Multiples  of 3

3 6 9 12 15  18  21  24  27  30  33  36  39  42  45  48  51   54  57 60  63  66  69  72  75

78  81   84  87   90   93  96  99

So

Multiples  of  3  that  are not also multiples of  4  or 5

3 6 18  21   27  33   39   42   51   54    57    63  66   69    78   81   87   93   99

So  multiples  of 2 or 3  that are not also  multiples of 4 or 5  are

2 3  6  9  14   18  21   22   26   27   33  34  38  39  42  46  51  54 57  58  62  63  66  69 74 78 81 82 86 87

93 94 98  99

So....34  ( if I counted  correctly)

Yes, your answer is right. The kite has no rotational symmetry. A kite also resembles a triangle reflected over one of its sides, forming a quadrilateral. Notice Δ ABD being reflected over its side from B to C, forming Δ CBD.

Say x is the number of truck owners.

Then the number of car owners would be x+6

Since all of Logan's coworkers either own a car or a truck, the equation would be x+x+6=20

which is equal to

2x+6=20

-6               -6

2x=14

x=7

7+6= 13 , so 13 people owned a car

Thanks !

X= 20, i know common core math is hard but come on dude

Thanks, heureka  !!!!

We  have  the form

qx^n  +   ....... + ax +   p

The possible rational roots   =   ± [  all  factors of   p    /   all  factors  of  q ]

Here   p  =  1      and  all  the possible  factors =  1

           q = 2       and all possible  factors =  1 , 2

So   all  the possible rational roors  =  ± 1  , ± 1/2   =    4  possible rational  roots

Find the value of the sum
\(\dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}\).

\(\text{Formula: } \qquad \dbinom{n}{k}= \dbinom{n}{n-k} \)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}} \\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{48} \\ && +\dbinom{99}{98}+ \dbinom{99}{96}+ \dbinom{99}{94} + \dots +\dbinom{99}{50}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4}+ \dbinom{99}{6} + \dots + \dbinom{99}{48} \\ && + \dbinom{99}{1}+ \dbinom{99}{3} + \dbinom{99}{5} + \dbinom{99}{7}+ \dots +\dbinom{99}{49}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{1} + \dbinom{99}{2}+ \dbinom{99}{3} + \dbinom{99}{4}+ \dots \\ && + \dbinom{99}{46} + \dbinom{99}{47} + \dbinom{99}{48}+\dbinom{99}{49}\\\\ &=& \dfrac{2^{99}}{2} \\\\ &=& \mathbf{ 2^{98} } \\ \hline \end{array}\)

Guest would be right if the groups were distinct, but if they are indistiguishable it would be half that or \(\frac{\binom{26}{13}}{2}\)

Just use RRT, you get the possible roots are \(\pm \frac12, \pm 1\) so there are 4 possible rational roots.

Very nice, Ilovepizza  !!!!

What does this mean, where is the problem?

We can see that if we put in all of the odd terms, we get \(2^{99}\) , however, the odd terms are missing. 

But, we see that for each even term \(\binom{99}{2x}\) there is a odd term \(\binom{99}{99-2x}\) which is equal to it.

So, the sum is \(2^{98}\).

Let   the  number  of   pens  =  3x

Let the  number  of erasers  =  x

Let  the  number  of  rulers  =  3x  - 28

And  we  know  that

x / ( 3x - 28)  =   3/2           cross-multiply

2x  =  3(3x - 28)         simplify

2x  =  9x  -  84        rearrange  

9x - 2x  =  84

7x  =  84

x =  84 / 7     =    12  =  number  of erasers

Number of  pens  =  3x   =  3(12)  =    36

Do you actually eat them or shove them up your dumb a?

You wouldn’t survive to.

As if I care.

Why should you care? You aren’t going to LIVE 20 years (MAYBE 20 minutes) if you eat 123456789 chocolates a day (and even doing that would be impossible!)

It is very unlikely that you will survive 20 days, never mind 20 years.

Way too much!

Maybe you should eat a little less…

A polynomial with integer coefficients is of the form

 \(2x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 1 = 0.\)
Find the number of different possible rational roots of this polynomial.

first is does not occur, second is a sample, third is a response.

The answers are 6, 8, 2, 2.

Mean is sum divided by number of elements so it is \(\frac{52}{7}\).

Median is simply 6

Mode is most often occuring so it is also 6

Range is 15-2=13

The three sides are (Letting area be K) \(\frac{2K}{3}, \frac{2K}{1}, \frac{2K}{h}\).

Then, simply do triangle inequality with cases depending on how big \(\frac{2K}{h}\) is.

\(29^{x/2} = \sqrt{29^x} = \sqrt{29}^x\) so \(A=\sqrt{29}\)

Minimum of -81 occurs at x=7,y=-13.

18 miles in 3 hours means 6 miles per hour upstream, 44 miles in 4 hours means 11 miles per hour downstream. So, let x and y be the speed of the boat and the current respectively, then x+y=11, x-y=6 so x=8.5, y=2.5

Case 1: m>0. Then, m<=8 so there are 8 values.

Case 2: m<0. Then, there are also 8 values, the negatives of the values in Case 1.

So, there 16 values total.