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He gave away 48 tarts.

If tan Z = 3/1 then cotan Z = 1/3

Same back to you MG.

I got immeasurable joy teaching you all those years ago. 

I still remember the massive effort you put in learning LaTex and even more so learning Combinatory mathematics. 

It is good to see you here too TR.

If it is parallel to the line y = 5x + 7, the line will have a formula of y = 5x + a

(7) = 5(-5) + a

Find a.

=^._.^=

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

this should be correct if not just let me know. hope this helps.

$\sqrt{\frac{3}{5}}=\frac{\sqrt{3}}{\sqrt{5}}$

$\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{3}}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}$

$\frac{\sqrt{3}}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{3*5}}{\sqrt{5*5}}$

$\frac{\sqrt{3*5}}{\sqrt{5*5}}=\frac{\sqrt{15}}{\sqrt{25}}$

$\frac{\sqrt{15}}{\sqrt{25}}=\frac{\sqrt{15}}{5}$

2x^2 - 3x + 27 = -4 has two imaginary roots. What is the sum of the squares of these roots?

Hello Guest!

$f(x)=2x^2 - 3x + 31 = 0$

$x = {-3 \pm \sqrt{3^2-4\cdot 2\cdot 31} \over 2\cdot 2}$

$x\in \{\frac{1}{4}(3+ i\sqrt{239}),\frac{1}{4}(3- i\sqrt{239})\}$

$\frac{1}{16}(3+ i\sqrt{239})^2+\frac{1}{16}(3- i\sqrt{239})^2=-\frac{115}{4}\color{blue}=-28.75$

  !

There are 3 pairs of congruent triangles.

I solved this problem before.  The correct options are B and C.

By Incusion-Exclusion, there are 7 venomous vermillion vipers.

< the acute angle between the hour and minute hands is 90 degrees >   

By the way, "acute angle" is defined as an angle that measures less than 90 degrees. 

Find all integers n such that when the clock strikes n o'clock, the acute angle between the hour and minute hands is 90 degrees. Enter your answer as a comma-separated list.     

When the clock striks the hour, then the minute hand is pointing at the 12.  

The positions that the hour hand can be in to make a 90^o angle are at the 3 or the 9.  

_.

what is 6x6 ?    

That's a wooden beam, but its actual dimensions are 5½ x 5½.  

Thx, I get it now 

y = k x^(1/4)

Let's start by finding k. 

3sqrt(2) = k * (81)^(1/4)

3sqrt(2) = k * 3

k = sqrt(2)

y = sqrt(2) * 4^(1/4)

Can you take it from here? :))

=^._.^=

Happy birthday Rosala. :))

=^._.^=

There are only 3 small primes:

2, 3, 5 ==2 + 3 = 5 ==10 -their sum

WOOOOO THE MATH GOD IS BACK.

just a couple more members left and this will be back to 2014/2015 times. glad to hear youre alive and doing well though, brother. 

2,020,202,020,201 mod 17 ==0

So, n==3

https://web2.0calc.com/questions/help-quadratic_20

3 o'clock and 9 o'clock

First GP:

4/12 ==1 /3 - this is the common ratio.

Sum==12 / [1 - 1/3]==12 / (2/3) ==18 - sum total of the 1st GP

Second GP:

Sum x 2 ==12 / [1 - R]

18 x 2 ==12 /[1 - R], solve for R

R ==2/3 - this is the common ratio of the 2nd GP

2nd term ==[4 + n]==12 x 2/3

[4 + n] ==8

n ==8 - 4 ==4