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Look at the 8.00 horizontal line...at which year does it intersect the graph, NSS ???

A bearing of 62.24° converts to an angle of 90 - 62.24 = 27.76°

And a bearing of  332.24°  converts to an angle of 332.24 - 270  = 62.24°

So we have a triangle  with angles of  27.76° and  62.24°

And the third angle of the triangle is  180 - 27.76 - 62.24  = 90°

So....we can find the distance from A to the coral reef using the Law of Sines

2590 / sin 90  =  D /  sin 62.24      where D is the distance we want

2590 / = D / sin 62.24       multiply both sides by sin62.24

2590 * sin 62.24  =  D ≈ 2292 ft.

Here's the approximate picture....dock "A" is a (0,0)....the reef is point B...and  dock "B"  is at point C

At least put this in an off-topic question.

Btw, it's probably C. I think.

To solve stem-and-leaf plots, just connect the number on the left to one of the numbers of the left. For the first row, it's 25, 26, and 27. You can figure it out from here.

I got the same answer (162). I too thought that rom doubled the number of games but now i see that his solution is correct, he just answered the question for 4 schools instead of 3.

If two lines in a plane are cut by a transversal so that a pair of alternate exterior angles is congruent, then the two lines are parallel.
In the image, if you know/prove ∠1 ≅ ∠8 OR ∠2 ≅ ∠7, then lines a and m are parallel.

See the image and this definition here:

https://quizlet.com/165214560/geometry-theorems-properties-etc-flash-cards/

Note: Scroll down about 3/4 of the way until you see " converse of alternate exterior angles postulate?"

Also, I think you doubled the answer, right? 4 students from School A competing with 4 students from School B is the same as 4 students from School B competing with 4 students from School A. Shouldn't the answer be 162?

Hey Rom, wouldn't it be 18 games against schoolmates? There are 3 schools not 4.

[Note: The number of terms of this expansion is calculated as follows:( 6(The exponent of the trinomial)) + number of terms - 1 nCr the exponent of the trinomial =6 + 3 -1 nCr 6 =8 nCr 6 =28 terms]

Thanks to my CAS system, here are ALL the 28 terms and their coefficients as calculated above:
4096 x^6 + 6144 x^5 y - 12288 x^5 z + 3840 x^4 y^2 - 15360 x^4 y z + 15360 x^4 z^2 + 1280 x^3 y^3 - 7680 x^3 y^2 z + 15360 x^3 y z^2 - 10240 x^3 z^3 + 240 x^2 y^4 - 1920 x^2 y^3 z + 5760 x^2 y^2 z^2 - 7680 x^2 y z^3 + 3840 x^2 z^4 + 24 x y^5 - 240 x y^4 z + 960 x y^3 z^2 - 1920 x y^2 z^3 + 1920 x y z^4 - 768 x z^5 + y^6 - 12 y^5 z + 60 y^4 z^2 - 160 y^3 z^3 + 240 y^2 z^4 - 192 y z^5 + 64 z^6
(28 terms)

Find the coefficient of x^2y^3z in the expansion of (4x + y - 2z)^6
The coefficient is calculated as follows:
[6!(original exponent) / 2!.3!.1!(exponents of the desired term)] X [4x(First term of original trinomial)]^(desired power(2)) X [y(Second term of original trinomial)]^(desired power(3)) X [2z(Third term of original trinomial)]^(desired power(1))
6! /2!.3!.1!  * (4x)^2 * (y^3) * (2z)=60 * 16x^2 * y^3 *2z=1920x^2y^3z

https://en.wikipedia.org/wiki/French_Republican_Calendar

Is this question supposed to make sense?

Vectors:

a)
Find CB:

\(\begin{array}{|rcll|} \hline -DC+DA+AB&=& CB \\ -p+2q-p+5p &=& CB \\ 2q + 3p &=& CB \\ \mathbf{\vec{CB}} & \mathbf{=} & \mathbf{2q + 3p} \\ \hline \end{array}\)

b)

\(\begin{array}{|rcll|} \hline AQ &=& \dfrac25 AB \quad & | \quad AB =5p \\ &=& \dfrac25 \cdot 5p \\ \mathbf{AQ} & \mathbf{=} & \mathbf{2p} \\\\ QB &=& \dfrac35 AB \quad & | \quad AB =5p \\ &=& \dfrac35 \cdot 5p \\ \mathbf{QB} & \mathbf{=} & \mathbf{3p} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{DA}{2}+PQ+QB-CB-DC &=& 0 \quad | \quad DA =2q-p \quad QB = 3p \quad DC = p \\ \dfrac{2q-p}{2}+PQ+3p-CB-p &=& 0 \\ \dfrac{2q-p}{2}+PQ+2p-CB &=& 0 \\ \dfrac{2q-p}{2}+PQ+2p &=& CB \\ && \mathbf{\vec{CB}=2q + 3p} \\ && 2q = CB - 3p \\ && q = \dfrac{CB - 3p}{2} \\ \dfrac{2\left(\dfrac{CB - 3p}{2}\right)-p}{2}+PQ+2p &=& CB \\ \dfrac{CB - 3p-p}{2}+PQ+2p &=& CB \\ \dfrac{CB - 4p}{2}+PQ+2p &=& CB \\ \dfrac{CB}{2} - 2p+PQ+2p &=& CB \\ \dfrac{CB}{2}+PQ&=& CB \\ PQ&=& CB -\dfrac{CB}{2} \\ \mathbf{PQ} & \mathbf{=} & \mathbf{\dfrac12\cdot CB} \\ \hline \end{array}\)

Vectors:

\(\begin{array}{|lrcll|} \hline (1) & a+b &=& RS \\\\ (2) & 2a+2b+BA &=& 2c \\ & BA &=& 2\left(c-(a+b)\right) \\\\ (3) & c+PQ - \dfrac{BA}{2} &=& 2(a+b) \\ &c+PQ- \dfrac{2\left(c-(a+b)\right)}{2} &=& 2(a+b) \\ &c+PQ- \left(c-(a+b) \right) &=& 2(a+b) \\ &c+PQ-c+(a+b) &=& 2(a+b) \\ &PQ +(a+b) &=& 2(a+b) \\ &PQ &=& 2(a+b)-(a+b) \\ &PQ &=& a+b \quad & | \quad a+b = RS \\ &\mathbf{PQ} & \mathbf{=} & \mathbf{RS} \\ \hline \end{array}\)

Are you asking for a value of a and b that will make this work?

\(\displaystyle \lim_{x^-\rightarrow1} \frac{\sqrt{x+a}\;+b}{(x^2-1)}=1\)

No idea.

A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a$, $b$, and $c$ are integers. Find $a+b+c$.

Formula:

\(\begin{array}{|rclcl|} \hline s_n &=& \dfrac{2}{3+1} + \dfrac{2^2}{3^2+1} + \dfrac{2^3}{3^{(2^2)}+1} + \dfrac{2^4}{3^{(2^3)}+1} + \dfrac{2^5}{3^{(2^4)}+1} + \ldots + \dfrac{2^{n+1}}{3^{(2^n)}+1} \\\\ s_0 &=& \dfrac12 \\ s_1 &=& \dfrac12 + \dfrac{4}{10} = \dfrac{9}{10} \\ s_2 &=& \dfrac{9}{10}+ \dfrac{8}{82}= \dfrac{409}{410} \\ s_3 &=& \dfrac{409}{410}+ \dfrac{16}{6562}= \dfrac{1345209}{1345210} \\ \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& 1 - \dfrac{a^{n+b}}{3^{ (2^{n+c} )}-1} \\\\ s_n &=& 1 - \dfrac{a^n a^b} { 3^{ (2^n 2^c )}-1 } \\\\ s_n &=& 1 - \dfrac{a^n a^b} { \left(3^{(2^c)} \right)^{2^n}-1 } \\ && \text{Set }~ \alpha=a^b \\ && \text{Set }~ \gamma=3^{(2^c)} \\ \mathbf{s_n} & \mathbf{=} & \mathbf{1 - \dfrac{a^n \alpha} { \gamma^{(2^n)}-1 } } \\ \hline \end{array}\)

\(\begin{array}{rcll} \boxed{n=0}:\\ s_0 = \dfrac12 &=& 1 - \dfrac{a^0 \alpha} { \gamma^{(2^0)}-1 } \\ \dfrac12 &=& 1 - \dfrac{\alpha} { \gamma-1 } \\ \dfrac{\alpha} { \gamma-1 } &=& 1 - \dfrac12 \\ \dfrac{\alpha} { \gamma-1 } &=& \dfrac12 \\ \mathbf{ \alpha } & \mathbf{=}& \mathbf{ \dfrac12\cdot (\gamma-1)} \qquad (1) \\ \end{array} \)

\(\begin{array}{rcll} \boxed{n=1}:\\ s_1 = \dfrac{9}{10} &=& 1 - \dfrac{a^1 \alpha} { \gamma^{(2^1)}-1 } \\ \dfrac{9}{10} &=& 1 - \dfrac{a \alpha} { \gamma^2-1 } \\ \dfrac{a \alpha} { \gamma^2-1 } &=& 1 - \dfrac{9}{10} \\ a\alpha & = & \dfrac{1}{10}\cdot (\gamma^2-1) \quad & | \quad \gamma^2-1 = (\gamma-1)(\gamma+1) \\ \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \\ \end{array}\)

\(\begin{array}{rcll} \boxed{n=2}:\\ s_2 = \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^{(2^2)}-1 } \\ \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^4-1 } \\ \dfrac{a^2 \alpha} { \gamma^4-1 } &=& 1 - \dfrac{409}{410} \\ a^2 \alpha &=& \dfrac{1}{410}(\gamma^4-1) \quad | \quad \gamma^4-1 = (\gamma^2-1)(\gamma^2+1)= (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\ \end{array}\)

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \quad & | \quad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \qquad (1) \\ a\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1) \\ a\cdot \dfrac12 & = & \dfrac{1}{10}\cdot (\gamma+1) \\ \mathbf{a} & \mathbf{=} & \mathbf{\dfrac{1}{5}\cdot (\gamma+1)} \qquad (4) \\ \hline \end{array} \)

\(\mathbf{\gamma=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\\\ && \mathbf{a = \dfrac{1}{5}\cdot (\gamma+1)} \quad (4) \qquad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \quad (1)\\\\ \dfrac{1}{25}\cdot (\gamma+1)^2\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \dfrac{1}{25}\cdot (\gamma+1)\cdot \dfrac12 & = & \dfrac{1}{410}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\gamma^2+\dfrac{5}{41} \\ \dfrac{5}{41}\gamma^2-\gamma-1+\dfrac{5}{41} &=& 0 \\ \dfrac{5}{41}\gamma^2-\gamma- \dfrac{36}{41} &=& 0 \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1-4\cdot\dfrac{5}{41}\cdot \left(-\dfrac{36}{41} \right) } } {2\cdot \dfrac{5}{41} } \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1+ \dfrac{720}{41^2} } } {\dfrac{10}{41} } \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \sqrt{1+ \dfrac{720}{41^2} } \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ \sqrt{2401} } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ 49 } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{41}{10} \cdot \dfrac{49} {41} \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{49}{10} \\\\ \gamma &=& \dfrac{41+49}{10} \\\\ \gamma &=& \dfrac{90}{10} \\ \mathbf{ \gamma }& \mathbf{=}& \mathbf{9} \\\\ \gamma &=& \dfrac{41-49}{10} \\\\ \gamma &=& -\dfrac{8}{10} \qquad \text{no solution, } \gamma\gt 0\\ \hline \end{array} \)

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline a & = & \dfrac15\cdot (\gamma+1) \quad & | \quad \gamma = 9 \\ a & = & \dfrac15\cdot (9+1) \\ a & = & \dfrac{10}{5} \\ \mathbf{a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{\alpha=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha & = & \dfrac12\cdot (\gamma-1) \quad & | \quad \gamma = 9 \\ \alpha & = & \dfrac12\cdot (9-1) \\ \alpha & = & \dfrac{8}{2} \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

\(\mathbf{b=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha &=& a^b \quad & | \quad a=2 \qquad \alpha = 4 \\ 4 &=& 2^b \\ 2^2 &=& 2^b \\ 2 &=& b \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{c=~ ?} \)

\(\begin{array}{|rcll|} \hline \gamma &=& 3^{(2^c)} \quad & | \quad \gamma = 9 \\ 9 &=& 3^{(2^c)} \\ 3^2 &=& 3^{(2^c)} \\ 2 &=& 2^c \\ 2^1 &=& 2^c \\ 1 &=& c \\ \mathbf{c} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{rcll} && \mathbf{a+b+c} \\ &=& 2+2+1 \\ &=& \mathbf{5} \\ \end{array}\)

Do you mean find the consecutive integers which span zero (i.e. one negative and the other positive)?

If so, the following table should help:

Hmmm!   

n        3^(n-1)(n-1)                n3^n

0        3^(-1)(-1) = -1/3         0*3^0 = 0

1        3^(1-1)(1-1) = 0         1*3^1 = 3

2        3^(2-1)(2-1) = 3         2*3^2 = 18

...etc.

1. A ball is thrown upward from the roof of a building 60 m tall. The ball reaches a height of 80 m above the ground after 2s and hits the ground 6 s after being thrown. Determine an equation of the function.

When t=0   h=60     (0,60)

when t=2   h=80       (2,80)

when t=6   h=0        (6,0)

The equation is    h=ax^2+bx+c

Use these points (substitution and simultaneous equations) to determine the values of a b and c.  

Well if you are in a hurry then that imples you are only really interested in the answer. 

Hence there is little reason for any of us to answer.

I am only interested in helping you to learn. I am not interested in providing you with a quick answer so that you can scrible it down, and then go off to do something  more fun.

Order of operations is a strict set of rules that dictate how to evaluate any expression. I have laid the order of operations out for you:
 

1. Simplify within grouping symbols such as parentheses or brackets from left to right.

2. Simplify exponents from left to right.

3. Perform multiplication or division, whichever operation comes first from left to right.

4. Perform addition or subtraction, whichever operation comes first from left to right.

If something is higher on the list, then it has greater priority. Let's apply this knowledge to this particular problem:

That final answer of weighing only 50kg after 2 days is so weird.

It does seem to be right though.

Originally 

weight = 100kg   water (99%) = 99kg

after 2 days

weight = 50kg    water (98%) = 49 kg

I'll take a look :)

3. Not as important ~ I have difficulty understanding this question: 
There was a flat containing boxes of apples having a total weight of 100 kg. An analysis showed that the apples were 99% moisture, by weight. After two days in the sun, a second analysis showed that the moisture content of the apples was only 98%, by weight. What was the total weight of the apples after 2 days, in kg?

\(\text {The first analysis shows the apples are 99% water. The weight of the water is then}\\ \left(0.99\cdot 100\right) = 99 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 100-0.98(100-x)\right)=x\\ \)

You want this last step explained. I shall try.  

there are 100kg of apples and originally 99% of this is water so there is  0.99*100=99kg of water originally in the apples.

Over time the water dries out. All the weight that the apples lose is because of the reduced water content. 

After 2 days the water contant is only 98% of the weight. All the rest of the apple is still there.

Now in that 2 days GingerAle  has let x represents the unknown weight LOSS of the apples.

So after 2 days the apples will weigh  (100-x) kg

98% of this will be water so the new weight will be  98% of (100-x) = 0.98(100-x)

Weight after 2 days = 0.98(100-x)

So the original weight was (0.99*100)kg and the new weight is  0.98(100-x) and the weight loss is x

so

it follows that

0.99*100  -   0.98(100-x)   = x

I hope that helps.

Xsquared's  answer is the correct one.     

You do need to bew careful though.

\(1)\;\;\;-1^2=-1*1^2=-1*1=-1\\~\\ 2)\;\;\;(-1)^2=-1*-1=+1=1\\~\\ 3)\;\;\; If \;\;x=-1\;\;then\;\; x^2=(-1)^2=+1\\~\\ 4)\;\;\; x^2=1\;\;find\;\;x \qquad \rightarrow \qquad x=\pm\sqrt1=\pm1\)

That is excellent :)