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The total cost of the meal would be 15 + tax + tip

tax = 7% of the meal = 7/100 x 15 = $1.05

tip = 18% of the meal = 18/100 x 15 = $2.70

total cost of meal = 15 + tax + tip = 15 + 1.05 + 2.70 = 18.75

By addition method, it is the same thing as elimination, which is what it is officially called.

Elimination:

So we can mulitply the bottom equation by 2 so that we can then cancle out the "x's" later on

(2x+3y=19)*2

4x+6y=38

Now we have the two equations

4x-y=3 , 4x+6y=38

We can then subtract these two equations to cancel out the "x"

4x-y-(4x+6y)=3-38

-y-6y=-35

-5y=-35

5y=35

y=7

So now we have figured out our y value is 7, and we can substitute it into our equation to get our "x" value now.

4x-7=3

4x=10

x=5/2 or 2.5 

Not so sure but here's what i think:

Since, Mars revolves around the sun, it travels at a rate of approximately 15 miles per second.
Since, 1 minute= 60 seconds. 
Therefore, 15 miles * 60 seconds= 900 miles/minute. (To make sure that this step is correct, Do as following: We know that velocity = Distance / Time in seconds.
So 900 (Distance) / 60 Seconds (Time in seconds) = 15 mile/sec.
 Which is the given so it is correct.
Since mars revolve around the sun and it travles at a rate of 900 miles per minute, 
So in 2 minutes it would revolve around the sun and it travels at a rate of 1800 miles per 2 minute.
Hope that helped.

∑[3^(1 - k) (2 k - 1) , k, 1, 1000] = 3

i am to

thanks for your answer, this is how I have done it now can you please correct me if I'm wrong? thanks

first I calculated the new volume of the cylinder with radius 10.03 and height 20.05;

v= pi * (10.03)² * ( 20.05) = 1263.56

The initial volume of the cylinder - the new volume =

6283.19 - 1263.56 = 5019.63

then I calculated the volume of the cylinder with radius 9.97 and height 19.95,

V= pi * (9.97)² * (19.95) = 6229.93

The initial volume of the cylinder - the new volume =

6283.19 - 6229.93 = 53.26 

so the approximate maximum error is 5019.63

Is that right? Please let me know. 

Yea ok, my logic was off.

My spreadsheet shows the truth of it :)

You could calculate the volume of a cylinder of radius 10+0.03 cm and height 20+0.05 cm, then determine the difference of the result from your previously calculated volume.

You should also do the same for radius 10-0.03 and height 20-0.05 cm.

Take the larger of the two differences.

I know 5 was your final answer, I just pointed out that to get that you divided 20 by 2 twice because you assumed that the number of ways to place 1-6 in the boxes such that they are ordered from left to right and such that the least number would be in the top left corner (ignoring columns being ordered from top to bottom) is 20, divided by 2 to find the number of such placements where the 2nd column is ordered and again to find the number of placements where both 2nd and third are ordered.

My final answer was 5, not 20.

I just worked out the answer manually.

My logic was a little off but the answer was still 5

Let us have four distinct collinear points A, B, C and D on the Cartesian plane.

The point C is such that AB/CB =1/2 and

the point D is such that DA/BA = 3 and

DB/BA = 2.

If C = (0, 4), D = (4, 0) and A = (x,y) what is the value of 2x+y?

\(\text{Let $AB=BA$} \\ \text{Let $DB=BD$} \\ \text{Let $\vec{C}=\dbinom{0}{4}$ } \\ \text{Let $\vec{D}=\dbinom{4}{0}$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{AB}{CB} &=& \dfrac{1}{2} \\ \mathbf{ CB } &=& \mathbf{2AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DB}{BA} &=& 2\\ \mathbf{ DB } &=& \mathbf{2AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline CD &=& CB + BD \\ CD &=& 2AB+2AB \\ \mathbf{ CD } &=& \mathbf{4AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DA}{BA} &=& 3 \\ \mathbf{ DA } &=& \mathbf{3AB} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DA}{CD} &=& \dfrac{3AB}{4AB} \\ \mathbf{\lambda_A=\dfrac{DA}{CD} }&=& \mathbf{\dfrac{3}{4}} \\ \hline \end{array}\)

Line:

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{A} &=& \vec{D}+\lambda_A(\vec{C}-\vec{D}) \quad | \quad \lambda_A = \dfrac{3}{4} \\ \vec{A} &=& \vec{D}+\dfrac{3}{4}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{3}{4}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{3}{4} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{3}{4}\cdot(-4)}{\dfrac{3}{4}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-3} {3} \\\\ &=& \dbinom{4-3}{3} \\\\ \mathbf{\vec{A} } &=& \mathbf{ \dbinom{1}{3} } \\\\ \vec{A} &=& \dbinom{x}{y} \qquad x = 1,\ y=3 \\ 2x+y &=& 2\cdot 1 + 3 \\ \mathbf{2x+y } &=& \mathbf{ 5 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DB}{CD} &=& \dfrac{2AB}{4AB} \\ \mathbf{\lambda_B=\dfrac{DB}{CD} }&=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

Line:

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{D}+\lambda(\vec{C}-\vec{D}) \\ \vec{B} &=& \vec{D}+\lambda_B(\vec{C}-\vec{D}) \quad | \quad \lambda_B = \dfrac{1}{2} \\ \vec{B} &=& \vec{D}+\dfrac{1}{2}(\vec{C}-\vec{D}) \\ &=& \dbinom{4}{0}+\dfrac{1}{2}\left( \dbinom{0}{4}-\dbinom{4}{0} \right) \\\\ &=& \dbinom{4}{0}+\dfrac{1}{2} \dbinom{-4}{4} \\\\ &=& \dbinom{4}{0}+ \dbinom{\dfrac{1}{2}\cdot(-4)}{\dfrac{1}{2}\cdot 4} \\\\ &=& \dbinom{4}{0}+ \dbinom{-2} {2} \\\\ &=& \dbinom{4-2}{2} \\\\ \mathbf{\vec{B} } &=& \mathbf{ \dbinom{2}{2} } \\ \hline \end{array}\)

a. Calculate the volume in cm^3 of figure A. 

b. Polyhedron B is a cylinder with on top of it a cone, what is the volume of polyhedron B in cm^3?

c. Polyhedron C is a block with a cylinder-like hole in it. The wood it is made of weighs 0,72 grams per cm^3. Calculate the weight of the block.

a. Show with a calculation that the height of the cone is 27cm

b. Calculate in litres the volume of the vase. 

Evaluate the sum

\(\large{s= 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb }\)

\(\begin{array}{|rcll|} \hline s &=& 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb \\\\ &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} + \dotsb \\\\ \hline s_n &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} \\\\ &=& 1\cdot 3^0 + 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{1-n} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& 1\cdot 3^0 + & 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{-n+1} \\ 3^{-1} s_n &=& & 1\cdot 3^{-1} + 3\cdot 3^{-2} + 5\cdot 3^{-3} + 7\cdot 3^{-4} + \dotsb + (2n-3)\cdot 3^{-n+1}+ (2n-1)\cdot 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} s_n-3^{-1} s_n &=& 1\cdot 3^0 +2\cdot ( 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}) - (2n-1)\cdot 3^{-n} \\ s_n(1-3^{-1}) &=& 1 +2\cdot ( \underbrace{3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}}_{\text{geometric progression}} ) - (2n-1)\cdot 3^{-n} \\ \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \\ \end{array}\\ \begin{array}{|rcll|} \hline S_n &=& 3^{-1} + & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} \\ 3^{-1} S_n &=& & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} + 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} S_n-3^{-1} S_n &=& 3^{-1} - 3^{-n} \\ S_n(1-3^{-1} ) &=& 3^{-1} - 3^{-n} \\ \dfrac{2}{3} S_n &=& 3^{-1} - 3^{-n} \\ \mathbf{ S_n } &=& \mathbf{\dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \quad | \quad \mathbf{ S_n = \dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\\\ \dfrac{2}{3} s_n &=& 1 +3\cdot ( 3^{-1} - 3^{-n} ) - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 +3^1 3^{-1} - 3^13^{-n} - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 + 1 - 3^{-n}(3 + 2n-1 ) \\\\ \dfrac{2}{3} s_n &=& 2- 3^{-n}(2n+2) \\\\ \dfrac{2}{3} s_n &=& 2- 2\cdot 3^{-n}(n+1) \\\\ \dfrac{2}{3} s_n &=& 2\Big(1- 3^{-n}(n+1)\Big) \quad | \quad : 2 \\\\ \dfrac{1}{3} s_n &=& 1- 3^{-n}(n+1) \\ s_n &=& 3\Big(1- 3^{-n}(n+1)\Big) \\ &=& 3\Big(1- \dfrac{n+1} {3^{n} }\Big) \\ &=& 3\Big(\dfrac{3^{n}-n-1} {3^{n} }\Big) \\ &=& \dfrac{3^{n}-n-1} {3^{-1}\cdot 3^{n} } \\ \mathbf{ s_n } &=& \mathbf{\dfrac{3^{n}-n-1} {3^{n-1} } } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s = \sum \limits_{n=1}^{\infty}\dfrac{2n-1}{3^{n-1}} = \lim \limits_{n\to\infty} s_n &=& \lim \limits_{n\to\infty}\dfrac{3^{n}-n-1} {3^{n-1} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n3^{n-1}-1} {(n-1)3^{n-2} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-3} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-2}3^{-1} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n} { (n-2)3^{-1} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{n} {n-2} \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {\dfrac{n-2}{n} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {1-\dfrac{2}{n} } \quad | \quad \lim \limits_{n\to\infty} \dfrac{2}{n} = 0 \\\\ &=& 3 \cdot \dfrac{1} {1-0 } \\\\ &=& 3 \cdot \dfrac{1} {1 } \\\\ \mathbf{ s } &=& \mathbf{3} \\ \hline \end{array}\)

Melody you are correct that the number of ways to choose 3 numbers out of 6 is 20 but the number of ways to place 1-6 in the boxes such that they are ordered from left to right AND such that the smallest number (1) is in the top left is not 20, it is 10.

Max continuity is not needed

EDIT: I don't think g(x) must pass through (0,0), f must pass through that point

g(x) will pass through (-3,-5)

I have not seen that symbol before either but 

8C5= nCr(8,5) = 56

How many ways can 3 numbers be chosen from 6?   6C3=20 ways

So there are 20 ways to chose two sets of 3 numbers.

Put each triplet in increasing order and put the one with she smallest number in the top left corner.

Now only half of those will have the second column in ascending order so that cuts it down to 10ways

But only half of those 10 will have the third culumn in ascending order so

There will be 5 possible outcomes.

Melody where in your proof did you use the fact that every room has an even number of doors?

^{Yes, but we are considering the house as a whole. Imagine a bird's eye view of the house with a transparent roof so that you can see all the doors.}

That is rubbish.  

The questions says, and I quote ^"every room has an even number of doors"

Each room is being considered independently of the other rooms.

It is not being viewed as  bird's eye view whatsoever.