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The angles could be  65  65  and x

OR

they could be   x, x,  and 65

The angle sume of a triangle is 180.

There are two possible answers. 

So how are you going to work them out?

PLEASE LET ASKER GUEST RESPOND. 

Here’s a neat and orderly way to transcribe this as a power of (2):

\(\hspace {15em} \huge2 \large^ {\log_2\left(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\right)}\\\)

GA 

--. .-

If x is replaced with x-3 then all the points will be moved in the positive x directions 3 places.

So translate the graph 3 units to the right.

What have you done for yourself?

This is a hint on how to start.

[(q+4)(p+3)]  +   {  -1 *    [(q+2)(p+1)]  } =44+2p+2q

Expand the bits in the square brackets first

Then do the braces curly brackets

Then look at what you have.

Try do do that and if you still need help post what you have tried.

Other answerers:  Please let guest asker attempt this. 

Do not do more for the asker unless he/she displays a proper attempt first

what is the value of y that is not attainable?

Hello Guest!

\(y=\frac{x-4}{4x-10}\ and\ \ x\neq 2\)

\(y=\frac{100000-4}{4\cdot 100000-10}=0.249996...\)

\(\frac{1}{4}\notin \{values\ of\ y\}\)

  !

Please check the question.

\(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\ \frac{4}{27}\left((2^{\frac{3+m}{3}})^3 + 3 (2^{\frac{3+m}{3}})^2 (2^{\frac{3}{m}}) +3(2^{\frac{3+m}{3}}) (2^{\frac{3}{m}})^2 + (2^{\frac{3}{m}})^3 \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(1+\frac{m}{3})})^2 (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(2+\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{2^2}{3^3}\left((2^32^m + 3 (2^2 2^{(\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2* 2^{(\frac{m}{3})}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \)

I can keep going but this is not going to fall out nicely.

LaTex:

\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\

\frac{4}{27}\left((2^{\frac{3+m}{3}})^3  +  3 (2^{\frac{3+m}{3}})^2  (2^{\frac{3}{m}})  +3(2^{\frac{3+m}{3}})  (2^{\frac{3}{m}})^2  +   (2^{\frac{3}{m}})^3 \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(1+\frac{m}{3})})^2  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(2+\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{2^2}{3^3}\left((2^32^m  +  3 (2^2 2^{(\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2* 2^{(\frac{m}{3})})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

[CFAH] = 1/3(AB * AD * DH)

EH = 5        NOT 15    your ###

There might be an easier way and you need to check my working for careless errors

\(z+\frac{1}{z}=1\\ \left(z+\frac{1}{z}\right)^2=1^2\\ \left(z^2+\frac{1}{z^2}+2\right)=1\\ \left(z^2+\frac{1}{z^2}\right)=-1\\ \left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\ \left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\ \left(z^3+\frac{1}{z^3}+1\right)=-1\\ \left(z^3+\frac{1}{z^3}\right)=-2\\ \left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\ \left(z^6+\frac{1}{z^6}+2\right)=4\\ \left(z^6+\frac{1}{z^6}\right)=2\\ \left(z^6+\frac{1}{z^6}\right)^2=2^2\\ \left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\ z^{12}+\frac{1}{z^{12}}=2\\ \)

LaTex:

z+\frac{1}{z}=1\\
\left(z+\frac{1}{z}\right)^2=1^2\\
\left(z^2+\frac{1}{z^2}+2\right)=1\\
\left(z^2+\frac{1}{z^2}\right)=-1\\
\left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\
\left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\
\left(z^3+\frac{1}{z^3}+1\right)=-1\\
\left(z^3+\frac{1}{z^3}\right)=-2\\
\left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\
\left(z^6+\frac{1}{z^6}+2\right)=4\\
\left(z^6+\frac{1}{z^6}\right)=2\\
\left(z^6+\frac{1}{z^6}\right)^2=2^2\\
\left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\
z^{12}+\frac{1}{z^{12}}=2\\

80 is the answer 

660/33 = 20 straws

20 * $.015 = $.30

K = C

(K-19)/(C-11) = 1/3

3K - 57 = C - 11

2K = 46

K = 23

C = 23

Completely simplify and rationalize the denominator.

Hello Guest!

\(\frac{\sqrt{160}}{\sqrt{252}}\times\frac{\sqrt{245}}{\sqrt{128}}\\ =\frac{\sqrt{2^5\cdot 5}}{\sqrt{2^2\cdot 3^2\cdot 7}}\times\frac{\sqrt{5\cdot 7^2}}{\sqrt{2^7}}\\ = \sqrt{\frac{2^5\cdot 5^2\cdot 7^2}{2^9\cdot 3^2\cdot 7}}\\ =\sqrt{\frac{5^2\cdot 7}{2^4\cdot 3^2}}\)

\(=\frac{5\cdot \sqrt{7}}{2^2\cdot 3} \)

  !

4x36=12²

\(\left(x^4+y^4\right)\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)\)

View Solution Here:

Let the number of pennies that Amanda has be 'p'.

It is given that she has one-half as many nickels as pennies

So, number of nickels = 1/2

It is given that she has three times as many dimes as pennies

So, number of dimes = 3 * 1/2 = 3p/2

Now, she has a total of 204 coins. So,

           p + p/2 + 3p/2 = 204

      => (2p + p + 3p)/2 = 204

      => 6p = 2 * 204

      => p = 408/6 = 68

So, Amanda has 68 pennies

So, number of nickels = p/2 = 68/2 = 34

and number of dimes = 3p/2 = (3 * 68)/2 = 102

Now, 1 penny = $0.01

         1 nickels = $0.05

    and 1 dimes = $0.10

So, the total money that Amanda has

    = $0.01 * 68 + $0.05 * 34 + $0.10 * 102

    = $0.68 + $17 + $10.2

    = $12.58

 Answer: $12.58 

Treat \(a^2\)as \(b \)

Now, when you factor \({b^2-3b+2}\)....

You get \((b-1)(b-2)\)

Remember, this equals \((a^2-1)(a^2-2)\)

If 8 numbers are chosen at random from the first 16 positive integers, what is the probability that an additional number chosen at random from the remaining 8 numbers is less than any of the 8 previously chosen numbers?

Not sure but this is my attempt

I'm thinking that maybe it doesn't matter that there are 16 numbers to start with.The fact is you have 9 numbers

The chance that the smallest is at the end is   1/9

This seems too simple to be correct but I cant see what is wrong with the logic.

-----------------------

If you find out the answer then please share it.   Preferably with working.

Thx for your help, but that was incorrect.

The way that you can get a product of 14 is:  1 x 1 x 4 x 6      12 ways

                                                                  or   1 x 2 x 2 x 6      12 ways

                                                                  or   1 x 2 x 3 x 4       24 ways

                                                                  or   2 x 2 x 2 x 3         4 ways

The number of possible ways is  1296.

Can you find the probability from these parts?

With   a = 8   b = 25   and   c = c

there will be two real roots when  b² - 4ac > 0.

b² - 4ac  >  0     --->     25² - 4·8·c  >  0     --->     625 - 32c  >  0

                                                                                     -32c  >  - 625   

                                                                                          c  <  625/32   =   19.53  (approx)

All integers in the range  1, 2, 3, ..., 17, 17, 19  will work.

Find their product ...