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By Inclusion-Exclusion, there are 6 student who do all three activities.

The parabola is 3x^2 - 2x + 8.

corrections: the equation in blue ink is incorrect!!! 

sqrt(4² - x²) = sqrt[17 - (6 - x)²]        x = 35/12

(AB)² = (2x)²

civonamzuk 

This is presented as a test...

No that's not right measure of arc is given in degrees or radian but that doesn't mean you can simply subtract it from 360 to calculate the length of larger arc. The formula for length of arc is 

                    length of arc $= \theta × {\pi r\over 180}$

where $\theta$ is the angle subtended by respective arc at the centre and r is the radius. 

So, we need the radius to calculate..

The circle centered at (2,-1) and with radius 4 intersects the circle centered at (2,5) and with radius sqrt(17) at two points A and B.  Find (AB)².

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Center C = (2, -1)         center D = (2, 5)           CD = 6

AC = 4          AD = √17

AB and CD intersect at point N           

CN = x          DN = 6 - x

x / 4 = (6 - x) / √17          x = - 24(4 - √17) = 2.954535015

(AB)² = (2x)²

If x, y, w, z are real numbers satisfying
 
$w+x+y = -2 \\ w+x+z = 4 \\ w+y+z = 19 \\ x+y+z = 12$

What is $wx + yz$?

$\begin{array}{|lrcll|} \hline (1)& w+x+y &=& -2 \\ (2)& w+x+z &=& 4 \\ (3)& w+y+z &=& 19 \\ (4)& x+y+z &=& 12 \\ \hline (1)+(2)+(3)+(4): & 3w+3x+3y+3z &=& -2+4+19+12 \\ & 3(w+x+y+z) &=& 33 \quad | \quad : 3 \\ & \mathbf{w+x+y+z} &=& \mathbf{11} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad x+y+z=12 \\ w+12 &=& 11 \\ w &=& 11-12 \\ \mathbf{w} &=& \mathbf{-1} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+y+z=19 \\ x+19 &=& 11 \\ x &=& 11-19 \\ \mathbf{x} &=& \mathbf{-8} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+z=4 \\ y+4 &=& 11 \\ y &=& 11-4 \\ \mathbf{y} &=& \mathbf{7} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+y=-2 \\ z-2 &=& 11 \\ z &=& 11+2 \\ \mathbf{z} &=& \mathbf{13} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline wx + yz &=& (-1)(-8)+7*13 \\ &=& 8+91 \\ \mathbf{wx + yz} &=& \mathbf{99} \\ \hline \end{array}$

p = 278.

First digit   4 choices

Second digit 4 choices

Third digt   2 choices   (only  3 and 5 as we want odd numbers)

4 x 4 x 2   =  32 possibles

x-coordinate of G = 3/2a

What was your old account in 2016?

https://web2.0calc.com/questions/help-plz-thx_2#r1

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The amount of mcc in each solution is the same...

.03 * 72   =  .02 (72+x)

2.16 = 1.44+ .02x

.72 =.02x

x =36 ml  added to 72 ml of  3%  to dilute it to 2%

The minimal possible value of |a - b| is 3.

The integer root is x = 2.

Ridiculous question... 

You could say 2 1/4 = 3/8p

2 1/4 is the total height of the stack, which is equal to 3/8 (thickness of each piece of paper) multiplied by p (placeholder for the total pieces of paper).

By solving for p, we are solving for the number of pieces of paper that will give us a stacked height of 2 1/4 inches.

We can solve this equation by dividing both sides by 3/8, giving us p = 6

Hope this helps!

~ シエラ

Sorry to hear this! Web2.0 calc is a free service, so it won't be as good as paid services such as Chegg or Study.com. From what I've seen, the mods and community here try their best to answer questions as quickly as possible! I've been around here roughly since 2016 (lost my other account, created this one recently), so feel to reach out to me, or one of the moderators!

We can use the Binomial Theorem here! (although you can also use the Pascal's Triangle method, I just prefer to use this)

(6!/(2!6-2)!)(2x)^4(1/2)^2 = 15*16x*1/4 = 60x^4

So therefore our coefficient is 60!

Hope this helps!

Sorry for this. The answer is actually $99$.Try adding all the equations together.

The solution is w = 1, x = -9, y = 6, z = 12, so wx + yz = (1)(-9) + (6)(12) = 63.

h doesn't change     (but k does)

2sqrt(17)

I hope this helps!!