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Area= Width x Length

A= 20*85=1700 

Perimeter= 2 (length + width)

P= 2(85+20)=210

thank you so much!

Yay lol

HAHAHA!!!....no need to doubt yourself, RP...you are completely  CORRECT  !!!!

The degree is found by adding the exponents of each term  and selecting the greatest sum

Note that we  could write this as :

5a^6b^1c^2d^0e^0f^0g^0h^0  + 8a^0b^0c^0d^5e^0f^0g^0h^0   + 7a^0b^0c^0d^0e^6f^2 - 10a^0b^0c^0d^0e^0f^0g^4h^7

So 

First term  = degree 9

Second term = degree 5

Third term = degree 8

Fourth term  = degree 11

So....degree 11

I'm probably completely wrong but at least i'm trying. Wouldn't you multiply 325.78 by 7.65%= $24.92 (D)

yes thanks for explaining!

A polynomial cannot  fractional exponents associated with any variable.....so the second one  is out

Also....

(30y^2 ) / x  =   30y^2 x^-1....and a polynomial cannot have negative exponents associated wiith any variable

So....

The first and last answers are  correct...[ only non-negative  non-fractional exponents are allowed ]

30y² + x    This is a polynomial.

30y² + x^1/2    If this your expression, it is not a polynomial because the exponent of x is a fraction.

(30y²)/x    This is not a polynomial because  x  is in the denominator.

30y²    This is a polynomial.

Even though he  contributes  8%  of his salary to his 401k....the company only matches up to 6%...so....

29512  x  .06  =  $1770.72  ⇒  "B"

Ah...an income tax question....don't get many of these  !!!

Since Elaine makes $520 /wk....we want to look at the row   where Column A  = 500  and Column B  = 600

Because    520 > 500  but  520 < 600

Now....since she has 4 exemptions,  on this 500- 600 row,  go over to the column under "4"  and you will find   the number "2"

I assume that this means  $2  withheld per week  ⇒  "B"

Do you see how to do this, NSS  ???

550 dollars per year /  12 months per year

=   550/12 dollars per month

≈   45.83 dollars per month

Width  = 80 - 21  =  59 inches

Perimeter = 2  ( Width + Length )  =

2 ( 59 + 80)  =  2(139)   =  278 inches

Area  =  W x L   =  59 x 80   = 4720 in^2

Here is a graph.

Can you work it out from that ?

Solving it this way gives:

\(a(a-1)^{-1} \equiv 4a^{-1} \pmod{20}\\ a \equiv 4a^{-1}(a-1) \pmod{20}\\ a^2 \equiv 4(a-1) \pmod{20}\\ a^2-4a+4 \equiv 0 \pmod{20}\\ (a-2)^2 \equiv 0 \pmod{20} \\ \text { }\\ a= 2+10n \;|\text { for n = (0, 1, 2, … 9), so 10 integers  satisfy 0 < a < 100}\)

Well blimey! You’re right. (I’m surprised because Heureka can do mod problems while napping.)

GA

There are no way to change your username ,once you enter then there are no optional way..!

A store would like to set up fish tanks that contain equal numbers of angle fish, sword fish, and guppies.

What is the greatest number of tanks that can be set up if the store has 12 angle fish, 24 sword fish, and 30 guppies?

Formula:

\(\begin{array}{|rcll|} \hline \gcd(a,b,c) &=& \gcd( \gcd(a,b), c ) \\ \gcd(a,b ) &=& \gcd (a-b,b ) & a \ge b \\ \gcd(a,a) &=& a & a \ge 0 \\ \gcd(a,b) &=& \gcd(b,a) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ && \gcd(30,24) \\ && = \gcd(30-24,24) \\ && = \gcd(6,24) \\ && = \gcd(24,6) \\ && = \gcd(24-6,6) \\ && = \gcd(18,6) \\ && = \gcd(18-6,6) \\ && = \gcd(12,6) \\ && = \gcd(12-6,6) \\ && = \gcd(6,6) \\ && = 6 \\\\ \gcd(30,24,12) &=& \gcd( \gcd(30,24), 12 ) \\\\ &=& \gcd( 6, 12 ) \\ &=& \gcd( 12, 6 ) \\ &=& \gcd( 12-6, 6 ) \\ &=& \gcd( 6, 6 ) \\ &=& 6 \\ \hline \end{array}\)

The greatest common divisor of two integers is \((x+2)\)
and their least common multiple is \(x(x+2)\),
where \(x\) is a positive integer.

If one of the integers is \(24\),
what is the smallest possible value of the other one ?

Let \(b\) is the value of the other one

greatest common divisor\((24,b) = x+2 \text{ and } x \in N\)

least common multiple\((24,b) = x(x+2) \text{ and } x \in N\)

Formula:

\(\begin{array}{|rcll|} \hline \text{greatest common divisor}(a,b) \times \text{least common multiple}(a,b) = a\times b \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline 24\times b &=& (x+2)\times x(x+2) \\\\ \mathbf{b} & \mathbf{=} & \mathbf{\dfrac{x(x+2)^2}{24}} \qquad x \in N, ~ x \gt 0 \\ \hline \end{array} \)

\(\begin{array}{|r|c|c|} \hline x \in N & \mathbf{b=\dfrac{x(x+2)^2}{24}} & \text{integer}\ ? \\ \hline 1 & \dfrac{3}{8} \\\\ 2 & \dfrac{4}{3} \\\\ 3 & \dfrac{25}{8} \\\\ 4 & \color{red}6 & \checkmark \\\\ \cdots & \cdots \\\\ \hline \end{array}\)

The smallest possible value of the other one is \(\mathbf{6}\) and \(x = 4\)

Check:

\(\begin{array}{lcl} \text{greatest common divisor}(24,6) &=& 4+2 = 6 \\ \text{least common multiple}(24,6) &=& 4(4+2)= 4\cdot 6 = 24 \\ \end{array}\)

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

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

If \(h(x) = g(f(x))\) 8 times, then:

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-8\quad & | \quad x = 12 \\ &=& 12-8 \\ &=& 4 \\ \hline \end{array} \)

Find

\(\large{\left(\dfrac{1+i}{\sqrt{2}}\right)^{46}}\)

\left(\frac{1+i}{\sqrt{2}}\right)^{46}.

\(\begin{array}{|rcll|} \hline && \left(\dfrac{1+i}{\sqrt{2}}\right)^{46} \\\\ &=& \dfrac{ \left(1+i \right)^{46} }{ \left(\sqrt{2} \right)^{46} } \\\\ &=& \dfrac{ \left( \left(1+i \right)^{2}\right)^{23} }{ 2^{\frac{46}{2}} } \quad & | \quad \left(1+i \right)^{2} = 2i \\\\ &=& \dfrac{ \left( 2i \right)^{23} } { 2^{23} } \\\\ &=& \dfrac{ 2^{23}i^{23} } { 2^{23} } \\\\ &=& i^{23} \\\\ &=& \left(i^{2}\right)^{11}i \quad & | \quad i^2 = -1 \\\\ &=& \left(-1 \right)^{11}i \\\\ &\mathbf{=}& \mathbf{-i} \\ \hline \end{array}\)

4x + 5a + 4y + 5b

                                   Numbers can be added in any order, so we can rearrange the expression.

=  4x + 4y + 5a + 5b

                                   Now we can factor  4  out of the first two terms and  5  out of the last two terms.

=  4(x + y) + 5(a + b)   Notice that if we distributed the  4  and the  5 , we would get the last expression.

                                   Since  x + y  =  -2 ,  we can substitute  -2  in for  x + y .

=  4( -2 ) + 5(a + b)

                                   Since  a + b  =  -2 ,  we can substitute  -2  in for  a + b .

=  4( -2 ) + 5( -2 )

=  -8  +  -10

=  -18

The smallest value of the other integer appears to be 60.

GCD[24, 60] =12, so (x + 2) =12 and x=10

LCM[24, 60] =120, so that x(x + 2) =10 x 12 = 120 

Find   \(\left(\frac{1+i}{\sqrt{2}}\right)^{46}\) .

Let's convert  1 + i  to polar form so we can use de Moivre's Theorem.

\(1+i=\sqrt2(\cos\frac{\pi}4+i\sin\frac{\pi}4)\)

Now lets raise both sides of that equation to the power of  46 .

\((1+i)^{46}=[\sqrt2(\cos\frac{\pi}4+i\sin\frac{\pi}4)]^{46}\\~\\ (1+i)^{46}=(\sqrt2\,)^{46}(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}\)

And divide both sides by  \((\sqrt2\,)^{46}\)

\(\frac{(1+i)^{46}}{(\sqrt2\,)^{46}}=(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}\\~\\ \left(\frac{1+i}{\sqrt{2}}\right)^{46}=(\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}\)

De Moivre's Theorem says....

\((\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)\)    So....

\((\cos\frac{\pi}4+i\sin\frac{\pi}4)^{46}=\cos(\frac{46\pi}{4})+i\sin(\frac{46\pi}{4})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{46\pi}{4})+i\sin(\frac{46\pi}{4})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{23\pi}{2})+i\sin(\frac{23\pi}{2}) \)

Now let's find a reference angle that is coterminal with  \(\frac{23\pi}{2}\)  by subtracting  \(2\pi\)  five times.

\(\frac{23\pi}{2}-2\pi-2\pi-2\pi-2\pi-2\pi\,=\,\frac{23\pi}{2}-10\pi\,=\,\frac{23\pi}{2}-\frac{20\pi}{2}\,=\,\frac{3\pi}{2}\)

So....

\(\left(\frac{1+i}{\sqrt2}\right)^{46}=\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=0+i(-1)\\~\\ \left(\frac{1+i}{\sqrt2}\right)^{46}=-i \)

In \(\left(\frac{1+i}{\sqrt{2}}\right)^{46}\) , we can apply exponent rule, \((\frac{a}{b})^c=\frac{a^c}{b^c}\) . If we do it here, we will get: \(\frac{(1+i)^46}{2^{23}}\) .\(\)