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It has 3 strips...and order matters

9 x 4 = 36?

Hello?

8 if you dont count the square ABCD and 9 if you count it 

Don't even expect CPhill to resort to you.... Shreyonica. You're not curious to know about anything.... except for Rohan because he's your BFF🤣🤣.

Shreyas, stop getting your answers for your homework through here.... That's only my job😜.

8+5 = 13 combined ratios of red and white toys.

8 / 13 x 351 =216 red toys

5/ 13 x 351 =135 white toys.

Ok... I guess I'm being really stupid or something.

Everything below is in degrees.

\(\sin(x) = \cos(x -90)\\ \sin(x-30) = \cos(x-120)\\ \text{we also have }\cos(x) = \cos(-x) \\ \text{and }\cos(2x + 360k)=\cos(2x)\)

as we move forward from x=0 the first place the equation is satisfied is at

\(\cos(2x) = \cos(120-x)\\ 3x=120\\ x=40\)

the second place the equation is satisfied is at

\(\cos(2x-360)=\cos(120-x)\\ 2x-360 = 120-x\\ 3x = 480 \\ x=160\)

the third place the equation is satisfied is at

\(\cos(2x-360) = \cos(x-120)\\ 2x-360=x-120\\ x=240\)

and the final place less than 2pi is at

\(\cos(720-2x) = \cos(x-120)\\ 720-2x=x-120\\ 840=3x\\ x = 280\)

So the solutions, as Melody showed graphically, are 

x = 40, 160, 240, 280 degrees.

The x axis intercept occurs when y= 0  

   so substitute y= 0 into the equation to find this value:

       3x - 5(0) = 7

           3x = 7

               x = 7/3   =   2  1/3

Thanks guest. 1 is not a prime number but that is not central to your arguement.

I am afraid your second paragraph makes no sense to me at all.

Thanks for you input though :)

I think you mean c = -6z. Assuming this is so, then:

(t-a)(t-b)(t-c) = 0

t³ - (a + b + c)t² + (ab + bc + ac)t - abc = 0

a + b + c = 2

(ab + bc + ac)/3 = -41

abc/(3*5*(-6)) = 6

t³ - 2t² - 123t + 540 = 0

See if you can now do part b.

4x- (x+8) 

there are many ways that you can think of this but the answer will always be the same.

Consider 

10 - (3+5)

the answer is 10-8 = 2

which is the same as   10 - 3 - 5 = 2

so

4x - (8+x) = 4x - 8 -x     =  3x-8

OR

If there is a minus sign in front of a bracket you can get rid of the bracket and the minus sign by changing the sign of everything within th bracket. Minus signs become + and plus becomes -

4x - (8+x)

=4x - (+8+x)

=4x    -8-x

=3x-8

OR

Another way I like to look at it is that you are adding -1 times the bracket

so

4x - (x+8)

=4x+     (-1)(x+8)

=4x+        -1*x + -1*8

=4x+           -x   -   8

=3x - 8 

\(AB^2=25+16-40cos(180-\theta)\\ AB^2=41+40cos \theta\\ AC^2=41-40cos\theta\\ AB^2+AC^2=82\\ M=m=82\\ M-m=0 \)

Thank you Rom...much appreciated!!..have a great day..

Heureka,

Thank you so much for the answer....how daft of me, to multiply the fraction instead of subtracting it...gosh....thank you kindly..

\(5(a+2) - 6 \div a = \\ 5(a+2) - \left(\dfrac{6}{a}\right)\\\)

\(\left . 5(a+2)-\dfrac 6 a \right |_{a=3} = \\ 5(3+2) - \dfrac 6 3 = \\ 25 - 2 = \\ \Large 23\)

yes

as x goes to infinity (1/3)^x goes to 0 so (1/3)^x​ - 5 goes to -5

If Log(5) = a, and Log(2) = b, determine Log(125) + Log(8) in terms of a and b

\(\begin{array}{|rcll|} \hline \boxed{125 = 5^3 \\ 8 = 2^3} \\ && \log(125) + \log(8) \\ &=& \log(5^3) + \log(2^3) \quad & \quad \boxed{\log(a^b) = b\times\log(a)} \\ &=& 3\times \log(5) + 3\times \log(2) \quad & \quad \log(5) = a,~ \log(2)=b \\ &=& 3\times a + 3\times b \\ &\mathbf{=}& \mathbf{3\times (a + b)} \\ \hline \end{array}\)

make "p" the subject.
\(\displaystyle x={5 \over3}- \sqrt{2p-1}\)

\(\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}\)

Hello Melody,

gosh, you really have walked the extra mile!!!!..many many thanks!!!...I soooo wish I had just some of your knowledge!!!..once again, thank you..

Thank you very much guest,

i do apreciate your input...

Another method:

5^3 / 10^-2 = 5^3  *  10^2 = 125 x 100 = 12500

Obviously the question only wants the pairs that add to between 20 an 30

So

20<=21+12k<=30

-1<=12k<=9

-1/12 <=k <= 9/12

-1

K= 0

There is only the one answer.

Hi Chris 

Once you have one you can find all

7(9)-5(12)=3

7(9+5k)-5(12+7k)=3

See I have added 35k and then taken it away again.