hairyberry

Start by counting all the perfect powers of non perfect powers of 2, 3, 5, 6, 7, 10 that are below 100. How many perfect powers we have, how many numbers we need to add onto 100. Then we repeat this process with the numbers from 100 to the new number we get until we finally have no numbers left (it wont be so confusing later on)

First we look at multiples of 2. There is 2^0 (we will only be counting 1 once) 2^2, 2^3, all the way to 2^6 ( 64). There is 6 numbers

Then look at multiples of 3. There is  3^2, 3^3, 3^4 There is 3 numbers

we skip 4 because 2 already includes all of these

From here, 5, 6, 7, and 10 all only have 1 multiple. 

We find the total number of multiples under 100 by doing 6 + 3 + 4  = 13 

Adding 13 to 100 we get 113.

We find that there are no numbers between 100 and 113 that are perfect powers therefore our answer is, 113

Triangle ABC is a right triangle because it has it's hypotenuse on a diameter of a circle. (point C is backwards on my graph, but that doesnt affect the result because its a reflection of the original image)

Finding AB with pythagorean theorem, we get \(AB = \sqrt{4^2 + 8^2} = 4\sqrt{5}\).

We also Get ACM=MCB = 45 degrees 

Because interior angles of a circle are congruent, we get AMB = BAM = 45. Now we see that ABM is a right isoceles triangle. Also, CAMB is a cyclic quadrilateral. We can calculate AM and BM too. \(AM = BM = \frac{4\sqrt{5}}{\sqrt{2}} = 2\sqrt{10}\)

Because ACBM is a cyclic, Ptolemy's theorem can be applied. 

Set the length of CM as X

then \(4*2\sqrt{10}+8*2\sqrt{10}=x*4\sqrt{5}\) and solving the equation we get x = \(x = 6\sqrt{2}\)

Please only ask one question per post.

Number 1: Only A can be expressed in the form ka*x, the others cannot therefore only A is an exponential function.

Number 2: 

A) 3^(2y) = 3^5, so 2y =5 y= 5/2

B) 9^(3u+5) = 3^8, 3^(6u+10) = 3^8, 6u + 10 = 8, u = -1/3

C) 8^(4x-1) = 8^10x, 4x -1 = 10x, x= -1/6

Number 3:

A) Reflecting a point over the x axis on a graph, (x,y) becomes (x, -y). We can use this to find that reflecting a function over the x axis we get that f(x) becomes -f(x). So after reflecting a exponential function over the x axis y = ka^x becomes y=-ka^x which is an exponential function.

B) Reflecting over the y axis, a point on a graph, (x,y) becomes, (-x,y). We can use this to fine the rule that to reflect a function over the y axis we apply the rule f(x) becomes f(-x). So an exponential function, y=ka^x becomes y=ka^(-x), which is k(1/a)^x, which is also an exponential function.

C)  Like the solutions above, we first translate a point vertically so (x,y) becomes (x,y+h). Applying that to functions in general we see that y or f(x) moves up h, and x stays the same, so a function transformation up h would result in f(x) becoming f(x)+h. Applying this to exponetial functions, a tranlation up h would be y=ka^x + h which is an exponential function. 

D) If a point is translated horizontally by g, then (x,y) would become (x+g, y) . Applying this to this situation, f(x) = ka^x, then moving right by g units would be k(a-g)^x (remember its minus g, minus moves right, plus moves left), which, by out definition is also an exponential function.

First calculate all the possible cases which is \(\frac{6!}{2!2!2!}\)= 90 

Then calculate all the possible cases which this condition occurs. The first time you pull there is three possible ways to pull a matching pair, 3*1, The second time you pull there is 2 possible ways to get a matching pair, 2*!, after you do this, you have to pull a matching pair. So (3*2)/90 = 1/15 

You could also do this with choose, (3/(6 choose 2))*(2/4 choose 2) * (1/2 choose 2) which also gives 1/15 

Whn lines are perpendicular, their slopes mutltiply to -1 in slope intercept form

We have: 

y=2x+13

6y+ax =6

y= 2x+13

6y = -ax+6

y=2x+13

y=-(a/6)x+1

-(a/6) = -1/2

a=3

The coefficents of the expansion of a trinomial is a lot like what the binomial theorem states about the coefficients of a binomial. 

For example, if you want the coefficient of the a^3*b^2*c term of the expansion of (a+b+c)^6, it is the same as taking 3as, 2bs, and a c, and ordering them which is 6!/(3!*2!*1!). 

The thing is we don't know in this problem is how many y^2 or 2ys are multiplied together to create the y^4 term, so we can split this into 3 scenarios:

2 y^2 terms are multiplied together to get y^4

4 2y terms are multiplied together to get y^4

2 2ys and 1 y^2s are multiplied together to get y^4

The coefficent for the first case would be 6!/(2!*4!)*(-5)^4 = 15 * 625 = 9375

The coefficent for the second case would be 6!/(2!*4!)*2^4*(-5)^2 = 15*16*25 = 6000

The coefficent for the third case would be 6!/(2!*3!*1!)*2^2*(-5)^2 = 60*4*(-125) = -30000 

Adding all the possible cases together we get: -14625 is the coefficient of the y^4 term

The expected value of the number of sixes is all the number of cases for the number of sixes * probability to get that number of 6s, all added together. 

The probability to get three sixes is (1/6)^3=1/216 

The probability to get two sixes is (1/6)*(1/6)*(5/6)*3 = 15/216

The probability to get one six is (1/6)*(5/6)*(5/6)*3 = 75/216

The expected value of sixes is 3*1/216 + 2*15/216 + 1*75/216 = 3/216 + 30/216 + 75/216 = 108/216 = 1/2

We have:

\(3x^2+9x+15=2x^2+8x-17\)

Start off by simplifying.

\(x^2+x+32=0\)

We want to find (r+1)(s+1) = rs+r+s+1

We apply Vieta's formula:

rs = 32

r+s = -1

32-1+1 = 32

First we use Stars and bars to find all possible cases:

\({x}_{1}+{x}_{2}+{x}_{3}...+{x}_{7}=9\)

Stars and bars tells us that there are (9+7-1) choose (7-1) = 5005 nonnegative integer solutions to this.

Now we assume Dhruv (assume that he is x₇) gets 2 chocolates. We can find the number of cases which Dhruv gets 2 chocolates. 

\({x}_{1}+{x}_{2}...+{x}_{6} = 7\)

Now we use stars and bars again to find that there are (7+6-1) choose (6-1) = 792 nonnegative integer solutions.

So the probability that Dhruv gets 2 chocolates is (792/5005) = (72/455)

(There might be a calculation error if there is tell me)