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Since one angle is  50°, then   x  = [ 180 - 50  ]  / 2  =   130  / 2   = 65°

Since two angles  = x, then these angles are equal

The triangle is

(a) acute  -  all angle are  < 90°

(b) isosceles -  two angles are equal, therefore the two sides opposite them are equal

Second one

y <  (-3/4) x + 1

This is a  dashed line - because of the "<"  - with a negative slope  and a y  intercept of  1

y > (2/3)x - 3

This  is also a dashed line with a positive slope and a y intercept of -3

Note that the graph   on the botom right is the only one satisfying these conditions....so, by default...it must be the  correct one

First one 

2x + y   ≥ -4  ⇒  y ≥ -2x - 4

This line will have a negative slope  and a y intercept of -4   .it will be   solid line because of the  " ≥"

y ≥ 2x

This line will also be solid...it will have a positive slope and a y intercept of 0

y - 3  > (1/3) x  ⇒  y  > (1/3)x  + 3

This will be a dashed  line  with a positive slope and  a  y intercept  of 3

We can eliminate the top left graph and the bottom right graph as these have lines with all positive sopes

Let's pick a point  ...say (-1, -1)  and see which equations this sattisfies.....notice that this point makes all three equations false...

So...the correct graph must be the one at the top right since the shaded area of this graph graph does not  include the point (-1, -1)

5% / 1200 =0.004166667 - monthly interest rate.

2,000 /1,000 =1.004166667^n, where n = number of months to double the money

2 = 1.004166667^n   take the log of both sides

n =Log(2) / Log(1.004166667)

n =166.7 months to double the money.Or:166 months - the least integer number of months.

n = 167 months when Kimberly will owe: $2,002.48

Note: I have taken your interest rate to mean "5% annual rate compounded monthly." Your question says "5% compounded monthly", which is equivalent to 79.59% effective annual rate.

If you DO MEAN 5% compounded monthly, then the above calculation will look like this:

2,000/1,000 =1.05^n

2 = 1.05^n

n = Log(2) / Log(1.05)

n =14.21 months, or: 14 - the least integer number of months, after which:

n = 15 months when Kimberly will owe $2,078.93

Nice solution! But, the answer is different help!

The operation \(*\) is defined for non-zero a andb  as follows: 

\(a * b = \frac{1}{a} + \frac{1}{b}. \)

If \(a+b=13\) and \(a \times b=25\), 

what is the value of  \(a*b\)  ?

Express your answer as a common fraction.

\(\begin{array}{|rcll|} \hline a+b &=& 13 \\ a\times b &=& 25 \\\\ \dfrac{a+b}{a\times b}&=& \dfrac{13}{25} \\\\ \dfrac{a}{a\times b} + \dfrac{b}{a\times b} &=& \dfrac{13}{25} \\\\ \dfrac{1}{b} + \dfrac{1}{a} &=& \dfrac{13}{25} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{13}{25} \quad & | \quad \dfrac{1}{a} + \dfrac{1}{b} = a*b \\\\ \mathbf{a*b} & \mathbf{=} & \mathbf{\dfrac{13}{25}} \\ \hline \end{array}\)

Modulo inverses
Find
\(24^{-1} \pmod{11^2}\).
That is, find the residue b  for which \(24b \equiv 1\pmod{11^2}\).
Express your answer as an integer from 0 to , inclusive.

\(\begin{array}{|rcll|} \hline & 24\cdot \underbrace{24^{-1}}_{=b} \equiv 1 \pmod{11^2} \\ \hline \end{array}\)

\(\begin{array}{rcll} \text{Let} \\ & 24\cdot 5 = 120 \equiv -1 \pmod{11^2} \\ \end{array} \)

\(\begin{array}{llcll} \text{square this equation: } \\ & (24\cdot 5)\cdot(24\cdot 5) &\equiv& (-1)\cdot(-1) \pmod{11^2} \\ & 24\cdot (5^2\cdot 24) &\equiv& 1 \pmod{11^2} \\ & 24\cdot 600 &\equiv& 1 \pmod{11^2} \quad | \quad 600 \equiv 116 \pmod{11^2} \\ & 24\cdot \underbrace{116}_{=(24)^{-1}=b} &\equiv& 1 \pmod{11^2} \\ \end{array} \)

\(\text{So $24^{-1} \pmod{11^2}=\boxed{116}$ is the multiplicative inverse to $24$ modulo $11^2$}.\)

Define
\(f(x)=\dfrac{1+x}{1-x}\)
and
\(g(x)=\dfrac{-2}{x+1}\).
Find the value of
\(\large{g(f(g(f(\ldots g(f(12)) \ldots ))))}\)
where there are 16 compositions of the functions g, and f,
alternating between the two.

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{x+1}{1-x} \\\\ g(f(x)) = \dfrac{-2}{\dfrac{x+1}{1-x}+1} &&=& x-1 \\\\ f(g(f(x))) = \dfrac{x-1+1}{1-(x-1)} &=& \dfrac{x+0}{2-x} \\\\ g(f(g(f(x)))) = \dfrac{-2}{\dfrac{x}{2-x}+1}& &=&x-2 \\\\ f(g(f(g(f(x))))) = \dfrac{x-2+1}{1-(x-2)} &=& \dfrac{x-1}{3-x} \\\\ g(f(g(f(g(f(x))))))=\dfrac{-2}{\dfrac{x-1}{3-x}+1} & &=& x-3 \\\\ f(g(f(g(f(g(f(x))))))) = \dfrac{x-3+1}{1-(x-3)} &=& \dfrac{x-2}{4-x} \\\\ g(f(g(f(g(f(g(f(x))))))))=\dfrac{-2}{\dfrac{x-2}{4-x}+1} & &=& x-4 \\\\ \ldots \\ \hline \end{array}\)

\(\text{When there are $\mathbf{16}$ compositions of the functions $g$, and $f$, alternating between the two.} \)

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-16 \quad & | \quad x = 12 \\ &=& 12-16 \\ &=& -4 \\ \hline \end{array}\)

 As we know that kelvin temperature scale is related to the degree celsius then we implement this by :

x=273+y

(a) -22 °C

273+(-22)=251 kelvin

(b) 0 °C

273+0=273 kelvin

(c) -32 °C

273+(-32)=241 kelvin

Hope that helped!

I have the following terms of an arithmetic sequence:

\(\frac{1}{2}, x-1, 3x, \ldots\)

$\frac{1}{2}, x-1, 3x, \ldots$.

Solve for \(x\).

Formula of an arithmetic sequence:

\(\begin{array}{|rcll|} \hline \begin{vmatrix} a_i & a_j & a_k \\ i & j & k \\ 1 & 1 & 1 & \\ \end{vmatrix} = 0 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array} \)

\(\text{Set $i=1$, $\ j=2$, $\ k=3$ } \\ \text{Set $a_i=a_1=\frac12$, $\ a_j=a_2=x-1$, $\ a_k=a_3=3x$ } \)

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ \frac12(2-3)+(x-1)(3-1)+3x(1-2) &=& 0 \\ \frac12(-1)+(x-1)(2)+3x(-1) &=& 0 \\ -\frac12+2x-2-3x &=& 0 \\ -\frac52-x &=& 0 \\ \frac52+x &=& 0 \\ x &=& -\frac52 \\ \mathbf{x} &\mathbf{=}& \mathbf{-2.5} \\ \hline \end{array} \)

I assume,

how to convert 82 mg to lb.

\(\begin{array}{|rcll|} \hline && \mathbf{83\ \text{mg}} \\\\ &=& 83\ \text{mg} \cdot \dfrac{ 1\ \text{kg} } { 1000000\ \text{mg} } \cdot \dfrac{ 2.2046226218\ \text{lb} } { 1\ \text{kg} } \\\\ &=& \dfrac{ 83\cdot 2.2046226218 } { 1000000 } \ \text{lb} \\\\ &\mathbf{=}& \mathbf{0.00018298368 \ \text{lb} } \\ \hline \end{array}\)

2.
Convert 373 milliseconds to units of megashakes.
One "shake" is equal to 10 nanoseconds[ns].

\(\begin{array}{|rcll|} \hline && \mathbf{373\ \text{ms}} \\\\ &=& 373\ \text{ms} \cdot \dfrac{ 1000000\ \text{ns} }{ 1\ \text{ms}} \cdot \dfrac{ 1 \ \text{shake} }{ 10\ \text{ns}} \cdot \dfrac{ 1 \ \text{megashake} }{ 1000000\ \text{shake}} \\\\ &=& \dfrac{ 373\cdot 1000000 }{ 10\cdot 1000000 }\ \text{megashake} \\\\ &=& \dfrac{ 373 }{ 10 }\ \text{megashake} \\\\ &\mathbf{=}& \mathbf{37.3\ \text{megashake}} \\ \hline \end{array} \)

1.

In Africa one "su" is 1.66 meters.

The average height of a human is 5 feet and 7 inches.

How to convert this height to units of "su" ?

feet to inches:

\(\begin{array}{|rcll|} \hline && 5\ \text{feet} + 7\ \text{inches} \\\\ &=& 5\ \text{feet} \cdot \dfrac{12\ \text{inches}}{1\ \text{feet}} + 7\ \text{inches} \\\\ &=& 60\ \text{inches} + 7\ \text{inches} \\\\ &=& 67\ \text{inches} \\\\ \hline \end{array} \)

inches to "su":

\(\begin{array}{|rcll|} \hline && \mathbf{67\ \text{inches}} \\\\ &=& 67\ \text{inches} \cdot \dfrac{ 2.54\ \text{cm} }{ 1\ \text{inch}} \cdot \dfrac{ 1\ \text{meter} }{ 100\ \text{cm}} \cdot \dfrac{ 1\ \text{su} }{ 1.66\ \text{meter}} \\\\ &=& \dfrac{67\cdot 2.54}{ 100\cdot 1.66} \ \text{su} \\\\ &=& \dfrac{170.18}{ 166} \ \text{su} \\\\ &\mathbf{=}& \mathbf{1.02518072289 \ \text{su}} \\ \hline \end{array}\)

THANK YOU ALL AND HAPPY B-DAY MAX

To make your fractions what you mean, you need to put parenthesees around each numerator and denominator.

Every fraction should be in this format:  ( stuff in numerator )/(  stuff in denominator  )

For example,     \(\frac{x^2-x-2}{x-3}\)    must be written with parenthesees like this:     (x^2 - x - 2)/(x - 3)

I can probably guess what your fractions are supposed to be, but since I already started making a point about this.....I don't want to just give it up.  

-3(y - 1)  =  9

                          First let's solve this equation for  y . Divide both sides by  -3 .

y - 1  =  9 / -3

y - 1  =  -3

                          Add  1  to both sides of the equation.

y  =  -3 + 1

y  =  -2

Now that we know  y  =  -2  ,   we can find the value of  \(\frac12\)y .

\(\frac12\)y  =  \(\frac12\)(-2)

\(\frac12\)y  =  -1

5280 ft / (2\(\frac12\) ft per stride)   =   5280 ft / (2.5 ft per stride)   =   2112 strides

So Jim must take  2112  strides to go exactly one mile.

num of strides Jim takes  =  2112

5280 ft / (2\(\frac23\) ft per stride)  =   5280 ft / (\(\frac83\) ft per stride)   =   1983.75 strides

So Jeffrey must take 1983.75 strides to go exactly one mile.

num of strides Jeffrey takes  =  1983.75

ratio of the num of strides Jim takes to the num of strides Jeffrey takes

=   num of strides Jim takes / num of strides Jeffrey takes

=   2112 / 1983.75

=   211200 / 198375

=   2816 / 2645

However...

if Jeffrey cannot take a fraction of a stride, he must take 1984 strides to reach a mile. If so...

num of strides Jeffrey takes  =  1984

And...

ratio of the num of strides Jim takes to the num of strides Jeffrey takes

=   num of strides Jim takes / num of strides Jeffrey takes

=   2112 / 1984

=   33 / 31

no thats not what i mean

Okay, thanks for editing it! But just to make sure.....

(3/2y) + (5/3y^2)   means   \((\frac32y)+(\frac53y^2)\)

(1/x-3) - (4/x-5)   means   \((\frac1x-3)-(\frac4x-5)\)

(x^2-x-2/x-3) times (x^2-9/x+1)   means   \((x^2-x-\frac2x-3)(x^2-\frac9x+1)\)

(3/p + q/6) / (2q/3p)   means   \(\frac{\frac3p+\frac{q}6}{2\frac{q}3p}\)

....Are these what you meant?

ok i did that

To make sure we understand your questions correctly, can you rewrite them to put parenthesees around each numerator and denominator?

For example.... to write the fraction  \(\frac{4a+1}{5b-2}\)  ,  you can write this:  (4a + 1)/(5b - 2)

The way it is currently written, this is your first expression:  \(\frac32y+\frac53y^2\)

But I think that might not be what you meant.

\(2a^4-a^3-6a^2-a+2\\ a^3(2a-1)-[(2a-1)(3a+2)]\\ (2a-1)(a^3-3a-2)\\ (2a-1)(a^3+1-3a-3)\\ (2a-1)[(a+1)(a^2-a+1)-3(a+1)]\\ (2a-1)(a+1)[(a^2-a+1)-3]\\ (2a-1)(a+1)(a^2-a-2)\\ (2a-1)(a+1)(a-2)(a+1)\\\)

I hope this helped,

Gavin. 

Consecutive terms of an arithemtic sequence have a common difference. 

We can write the equaiton:

\((x-1)-\frac12=3x-(x-1) \\ x-1.5=2x+1\\ x=-2.5\)

Then, we can plug this value into the terms to check if they have a common difference. 

The sequence goes: \(\frac12, -2.5-1, 3\cdot(-2.5) \Rightarrow0.5, -3.5, -7.5\)

The common difference is 4, so the value works!

I hope this helped,

Gavin. 

Isn't the gap between two consecutive numbers, the same? So, \(3x-x-1=\frac{1}{2}, 2x-1=\frac{1}{2}, 2x=\frac{3}{2}, x=\frac{3}{4}\) . 

We know that: \(\frac{1}{a}+\frac{1}{b}=a*b\) . If we multiply the denominator by b and a, respectively, we get: \(\frac{1b}{ab}+\frac{1a}{ba}=a*b\) . Suddenly, we go back to the problem, and it says: \(ab=25.\) So, we have:\(\frac{1a}{25}+\frac{1b}{25}=\frac{1a+1b}{25}.\) We also know that \(a+b=13\) , so  \(\frac{1a+1b}{25}=\frac{a+b}{25}=\boxed{\frac{13}{25}}.\)