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​Classify by degree and number of terms.

The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$.
Triangle $BIG$ is translated five units to the left and two units upward to triangle $BIG'$,
in such a way that
$B'$ is the image of $B$,
$I'$ is the image of $I$, and
$G'$ is the image of $G$.
What is the midpoint of segment $B'G'$? Express your answer as an ordered pair.

1.

\(\text{The midpoint of $BG$ is $\frac{B+G}{2}$ }\)

\(\begin{array}{|rcll|} \hline && \dfrac{B+G}{2} \\\\ &=& \dfrac{\dbinom{1}{1}+\dbinom{5}{1}}{2} \\\\ &=& \dfrac{\dbinom{1+5}{1+1}}{2} \\\\ &=& \dfrac{\dbinom{6}{2}}{2} \\\\ &=& \dbinom{\frac{6}{2}}{\frac{2}{2}}\\\\ &=& \dbinom{3}{1}\\ \hline \end{array}\)

The midpoint of BG is \(\mathbf{(3,\ 1)}\)

2.

\(\text{The translation of the midpoint of $BG$ is the midpoint of $B'G'$ }\)

\(\begin{array}{|ll|} \hline \text{Translation}\\ x' = x-5 \quad & \text{five units to the left} \\ y' = y+2 \quad & \text{two units upward} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \underbrace{\dbinom{3}{1}}_{\text{midpoint of $BG$ }} \rightarrow \quad \underbrace{\dbinom{3-5}{1+2}}_{\text{translation}} &=& \underbrace{ \mathbf{ \dbinom{-2}{3}} }_{ \text{midpoint of $B'G'$} } \\ \hline \end{array}\)

The midpoint of B'G' is \(\mathbf{(-2,\ 3)}\)

The LA=2sl where s is the length of the square base. So the lateral surface area us D 126.

thank you

Let us say everything is positive. Now the ratio of a 45-45-90 triangle is 1:1:\(\sqrt2\) . But the ratio on the question is  \(\frac{\sqrt3}{2}\)

Let the distance = D

Average speed = Distance travelled / Time

Time = Distance / speed

D/12  +  D/8 = 1 hour, solve for D

[2D  +  3D]  / 24 = 1 hour                 cross multiply

5D = 24                                                 divide both sides by 5

D = 24/5

D = 4.8 miles - the distance from her home to shchool.

The area of half the saded region is 

144pi/6-36sqrt3=29pi-36sqrt3

Then, double that is 58pi-36sqrt3.

Easy, rectangle

Let us have d= the distance from home to school. Then the time at 12mi/hr is 5d because 12/60=1/5. Similarily 8/60=2/15 so it takes 15/2*d. Adding the times gives us 12.5d=60 so d=4.8mi. Thus it takes 4.8mi to school.

In order to answer this question, one must determine what the solutions are to the equation \(18+5x^2=20x\). Let's do that, shall we?

By Pythagoras's Theorem:

25^2 - (14/2)^2 = AD^2

AD^2 = 576

AD = sqrt(576)

AD = 24

A period is the length from one peak to the next immediate peak on a periodic function.

In this particular problem, there is a peak at \(-\frac{3\pi}{4}\) ,and there is a peak at \(\frac{\pi}{4}\). The distance between the two is \(|\frac{-3\pi}{4}-\frac{\pi}{4}|\) or \(\pi \) units. 

Find the product of all positive integer values of c such that 3x^2+7x+c=0 has two real roots.

Hello Lightning!

\(3x^2+7x+c=0.\) 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-7 \pm \sqrt{7^2-4*3c} \over 2*3}\\ x ={ -7 \pm \sqrt{49-12c} \over 6}\)

There are only two real roots if 12c < 49.

\(c \in\{1,2,3,4\}\)

The product is 4! = 24

Greetings

  !

You seem to understand your transformations well. This looks right to me!

40 / 32 = 1.25 - 1 x 100 = 25% - perecentage increase.

\(\text{midline}=\frac{\text{max value}+\text{min value}}{2}\)

Since the amplitude of a sinusoid is equal to the vertical distance from its midline to one of its extrema, we can also calculate the midline using one of the following formulas.

\(\text{midline}=\text{max value}-\text{amplitude}\)

\(\text{midline}=\text{min value}+\text{amplitude}\)

I'm going to leave the calculating to you. 

I hope this helped,

Gavin

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\)

\(i=\sqrt{-1}, \text{ we start simple and try to find a pattern.}\)

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \)

Every 4 terms, the sequence rotates, \(1, i , -1, -i\)

Their sum is: \(1+i-1-i=0\)

Since every four terms cancel out, we just need to know what term the sequence starts and ends. 

\(i^{-100}=1\\ i^{100}=1\\\)

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\\ =1+i-1-i+1+-\cdots+1+i-1-i+1\\ =\boxed1\)

I hope this helped,

Gavin

Hey FiestyGeco!

I can only help you with 3, since the pictures won't show. 

For 3, we use \(c^2=a^2+b^2-2ab\cos\gamma\), the law of cosines.

This formula generalizes the Pythagorean Theorem, since \(\cos90º=0\Rightarrow a^2+b^2=c^2\)

We can write: \(CB^2=CA^2+AB^2-2(AC)(AB)-\cos{A}\)

\((4\sqrt3)^2=(4\sqrt2)^2+AB^2-2(4\sqrt2)(AB)-\cos60\\ 48=32+AB^2-8\sqrt2AB-\frac12\\ \)

After solving for AB, you use the same formula to solve for the angle.         

This doesn't look like the best method, does anyone know a less rigid way of doing this?

I hope this helped,

Gavin

If you can write a computerized algorithm to solve it, then you would have to at least have some idea how to solve it. 

Your question: Whats an alternative answer to brute forcing though? (sic)

The answer: Use an optimized partitioning algorithm.

Determine the sum of the missing scores by solving (10 + 10 + 10 + 9 + 9 + 7+ 7 + x)/10 =7.8. (x=24).

Find the partitions in set of three summing to 24. There are 48 of them; consider only the partitions that have values of 10 or less.

(This is easy enough to do without a computer, for this problem.)

10+10+4

10+9+5

10+8+6

10+7+7

9+9+6

9+8+7

8+8+8

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Calculate the medians:

{10, 10, 10, 9, 9, 7, 7, 7, 5, 4} Mean= 7.8. Median = 8.0

{10, 10, 9, 9, 9, 7, 7, 7, 5, 5} Mean= 7.8. Median = 8.0

10, 10, 9, 9, 8, 7, 7, 7, 6, 5} Mean= 7.8. Median = 7.5

10, 10, 9, 9, 7, 7, 7, 7, 7, 5} Mean= 7.8. Median = 7.0

10, 9, 9, 9, 9, 7, 7, 7, 6, 5} Mean= 7.8. Median = 8.0

10, 9, 9, 9, 8, 7, 7, 7, 7, 5} Mean= 7.8. Median = 8.0

10, 9, 9, 8, 8, 8, 7, 7, 7, 5} Mean= 7.8. Median = 8.0

Extract unique medians and calculate the sum (8+7.5+7) = 22.5

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I do wonder if StudyDaddy is scraping this forum.

GA

Can you provide a diagram plz.

Ok, I had no idea how to do this so I just wrote a python script. 

It appears the answer is 7+7.5+8=22.5

Whats an alternative answer to brute forcing though?

There are 10^3 - 10^2 = 900 3-digit numbers with repeats allowed. If we begin with 1, we see the following pattern:
101, 121, 131, 141, 151, 161, 171, 181, 191. As you can see, no 2 SAME  digits are adjacent to each other. This pattern holds for all remaining 8 numbers. Therefore, these numbers total: 9 x 9 = 81, which we must subtract from 900 above and we get:
900 - 81 = 819 3-digit numbers.