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Here is how to find the 'x' coordinate....(you try to find the 'y')

from endpoint x   to midpoint x    is   3   to  11      a distance of 8    add 8 more to the midpoint x to get the other endpoint 11+8 = 19 = x

  Now you try the same for the 'y' coordinate.....

Since RQT is a straight line it = 180 degrees

   so the two angles sum to 180

       9x+10     +  6x+5      =   180    

             15x     + 15      = 180          Now, you should be able to solve for 'x'....then RQS =   9x+10

As follows:

Halp this matters a LOT

Number Theory

There is a number formed by n copies of 2020 and 1 in the unit place.
If this number is divisible by 17,
what is the smallest possible value of n?

Formula: \(2020n+1 \equiv 0 \pmod{17} \)

\(\begin{array}{|rcll|} \hline 2020n+1 &\equiv& 0 \pmod{17} \quad &|\quad 2020 \equiv 14 \pmod{17} \\ 14n+1 &\equiv& 0 \pmod{17} \\ 14n &\equiv& -1 \pmod{17} \\ n &\equiv& (-1)\dfrac{1}{14} \pmod{17} \\ \hline \end{array}\)

Modular multiplicative inverse using Euler's theorem:

\(\begin{array}{|rcll|} \hline \dfrac{1}{14} \pmod{17} &\equiv& 14^{\phi(17)-1} \pmod{17} \quad &|\quad \phi(17)=16 \\ &\equiv& 14^{16-1} \pmod{17} \\ &\equiv& 14^{15} \pmod{17} \quad &|\quad 14 \equiv -3 \pmod{17} \\ &\equiv& (-3)^{15} \pmod{17} \\ &\equiv& -(3)^{15} \pmod{17} \\ &\equiv& -(3)^{5*3} \pmod{17} \\ &\equiv& -\left(3^5\right)^{3} \pmod{17}\quad &|\quad 3^5 \equiv 5 \pmod{17} \\ &\equiv& -(5)^3 \pmod{17} \\ &\equiv& -125 \pmod{17} \quad &|\quad 125 \pmod{17} \equiv 6 \pmod{17} \\ \mathbf{ \dfrac{1}{14} \pmod{17} } &\equiv& \mathbf{ -6 \pmod{17}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline n &\equiv& (-1)\dfrac{1}{14} \pmod{17} \quad &|\quad \mathbf{ \dfrac{1}{14} \pmod{17} \equiv -6 \pmod{17}}\\ n &\equiv& (-1)(-6) \pmod{17} \\ \mathbf{n} &\equiv& \mathbf{6 \pmod{17}} \\ n &=& 6+17k \qquad k\in \mathbb{Z} \\ \hline \end{array}\)

The smallest possible value of \(n\) is \(\mathbf{6}\)

x = total questions

questions answered = 5/8 x + 13

  and this is   6 out of (6+1) = 7    or  6/7 of the total

5/8 x + 13 = = 6/7 x

x = 56 questions

12 animals   *   1/2 clay / animal  =  6 clays

F = 43 234 ( 1 - .012)^t

n = 6 by trial and error .... (  2020 * 6   + 1  ) / 17 = 713

a)  v = 128    ho = o  (ground level)

     h(t) =  - 5t² + 128t    

b) h(t) will be zero when rocket hits the ground

      0 = -5t^2 + 128t        solve for t

c)  112 = -5t^2 + 128 t      use quadratic formula to solve for t 's     subtract smaller from larger value to find answer

d) maximum height will occcur at t = - b/2a    where b = 128 and a = -5    

          e)   use this value of 't' to calculate the maximum height

What (factor of 6 ) + 3    will leave 3 when divided by 4 ?

9    no

15  yes                 15 divided by 6 would leave 3       15 divided by 4 would also leave 3

12 inches = 1 foot

so the cage has an area of   L x W =  1  x  2.5 = 2.5 sq ft

2.5 sq ft / (2/3 sqft per hamster ) =  2.5  * 3/2  = 7.5 / 2 = 3.75 hamsters    Basically enough room for 3 Dwarf Hamsters

Can you help with this problem?

There is a number formed by n copies of 2020 and 1 in the unit place. If this number is divisible by 17, what is the smallest possible value of n?

Wouldn't this be 15 c 2 = 105? Does the savory or sweet part even matter?

p(x)*q(x) = g(x)^2-f(x)^2 = x^6-2x^3+1-x^2-10x-25 = x^6-2x^3-x^2-10x-24

h=4s so 5s = 20.6 which means s=4.12 and h = 4s=4*4.12 = 16.48

The six-digit integer 978XYZ consists of six distinct digits and is divisible by

7, 8 and 9.
What is the three-digit integer XYZ?

\(\begin{array}{|rcll|} \hline 7 | 978XYZ \\ 8 | 978XYZ \\ 9 | 978XYZ \\ \hline \end{array} \begin{array}{|rcll|} \hline XYZ =~ ? \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 7*8*9 &=& 504 \\ 978000 : 504 &=& 1940~ R~ 240 \\ 240+\mathbf{264} &=& 504 \\ \mathbf{XYZ} &=& \mathbf{264} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 978000 +264 &=& 978264 \\ 978264 : 7 &=& 139752 \\ 978264 : 8 &=& 122283 \\ 978264 : 9 &=& 108696 \\ \hline \end{array}\)

The inverse of a modulo 39 is b. What is the inverse of 14a modulo 39 in terms of b? Give your answer as an expression in terms of b.

I did this a really super long way using the Euclidean algorithm and got 

14a*14b=1 mod39

so the inverse of 14a is 14b  (mod39)

Maybe there is an easier way.  In this case trial and error would have been easier......

It is correct though   14a*14b= 196ab = 1*1 (mod39) = 1  (mod39)

Probably the fact that 39=3*13 should have been used, but I am not sure exactly how.  (I have never studied modulo arithmetic formally)

What is the product of the two smallest prime factors of \(2^{100} - 1\)?

First prime factor 2 ?
\(2^{100} - 1\) is an odd number so 2 is no prime factor in \(2^{100} - 1\)

next prime factor 3 ?
If 3 is a prime factor in \(2^{100} - 1\), then \(2^{100} - 1 \equiv 0 \pmod{3}\)
\(\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \quad | \quad 2 \equiv -1 \pmod{3} \\ (-1)^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \\ 1-1 &\equiv& 0 \pmod{3} \quad ? \\ 0&\equiv& 0 \pmod{3}~ \checkmark \quad | \quad {\color{red}3} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}\)

next prime factor 5 ?
If 5 is a prime factor oin \(2^{100} - 1\), then \(2^{100} - 1 \equiv 0 \pmod{5}\)
\(\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 2^{2*50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ \left(2^2\right)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \quad | \quad 2^2=4 \equiv -1 \pmod{5} \\ (-1)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 1-1 &\equiv& 0 \pmod{5} \quad ? \\ 0&\equiv& 0 \pmod{5} ~ \checkmark \quad | \quad {\color{red}5} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}\)

\(3*5 = 15\)

The product of the two smallest prime factors of \(2^{100} - 1\) is 15

Factor 1560 into its prime factors:

1560 = 2^3*3*5*13

To make this into a perfect square multiply by 2*3*5*13 = 390

n^2 = 1560*390 = 608400

Hence n = 780

Look at the coefficients of the powers of x.  These are 1, 10 and 27.  Their sum is  1+10+27 = 38.  Hence, if x = 1 then f(x) = 0, so the integer root is x = 1.

Just add them up, easy peazy