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It was Jack’s boo-boo. ...Jill kissed it, and made it all better. 

I assume you want the equation

The vertex is (-4,2)  and it passes though (-2,0)

\((y-2)^2=4a(x+4)\\ sub\;\; in \;\;(-2,0)\\ 4=4a*2\\ 2a=1\\ a=0.5\\~\\ (y-2)^2=2(x+4)\\\)

now rearrange to get it into the forem you want.

One of us obviously made a boo boo Ginger :)

1. In a paintball game, Jack attempted 20 shots and made 15 in the first half. In the second half, he was a marksman and made every single one of his 10 shots.

Jill scored a lower percentage of shots than Jack in each of the halves, but had exact same overall percentage as Jack.

If Jill attempted 12 shots and 18 shots in the first and second half respectively, how many baskets did Jill score in each?

Jack attempted 30 shots and made 25.      25/30 

Jill attempted 30 shots She scored the sam percentage in total  but less spercentage in each half

First half

Jack got 3/4

Jill got less than 3/4 of 12  so less than 9.    That would be a max of 8 out of 12

Second half

Jack got 100% so Jill go less than that, a maximum of  17 out of 18

\(\frac{8-c+17-k}{30}=\frac{25}{30}\\ 25-(c+k)=25\\ c+k=0\\\)

So Jill got 8 in the first half and 17 in the second half

Plug it in to a calculator, and you get \(11.8780382733 \)

Solution:

Jack’s fist-half percentage is 75% with 15 successes,

Jack’s second-half percentage is 100% with 10 successes

Jack’s total percentage is 83.333% with 25 successes in 30 shots.

Jill’s total percentage is also 83.333% with 25 successes in 30 shots.

Jill’s fist-half percentage has to be less than 75% for 12 shots.

So, 0.75*12 = 9. Jill had 8 successes in 12 attempts (66.666%).

Then 25 - 8 = 17, so Jill had 17 successes in 18 attempts (94.44%).

Each of these percentages is less than Jack’s percentages for the first and second half.

Jill had (8) successes in the first half and (17) successes in the second half.

GA

--. .-

\(2^{10} = 2^{10}\)

\(4^{20} = 2^{2^{20}} = 2^{40}\)

\(8^{40} = 2^{2^{2^{40}}} = 2^{160}\)

\(2^{10} * 2^{40} * 2^{160} = 2^{210}\)

So... \(n = 210\)

Jack and Jill went up the hill

To fetch a pail of water.

Jack fell down and broke his crown,

And Jill came tumbling after.

When the fun was done

Up Jack got and down he trot

As fast as he could caper;

And went to bed and covered his head

In vinegar and brown paper.

When Jill came in how she did grin

To see Jack's paper plaster;

Mother vexed, did whip her next,

For causing Jack's disaster.

Now Jack did laugh and Jill did cry

But her tears did soon abate;

Then Jill did say that they should play

At see-saw across the gate.

The next day, Jack and Jill when up the hill to shoot paintballs at a basket.

After they finished playing they solved this problem.

Solution below....

No explanation is needed.

EF = 7

This is the binary representation for 'correct'

01100011011011110111001001110010011001010110001101110100

Sum==[F  +  L] / 2  x  N, where F==First term, L==Last term, N==Number of terms.

The smallest multiple of 6 ==6. The largest multiple of 6 under 600==594

[594 - 6] / 6 + 1 ==98 + 1 ==99 terms.

Sum==[6  +  594] / 2  x  99 ==29,700. So, the units digit==0.

x_1 + x_2 = -a

x_1 x_2 = 1

x_1 + x_2 = -1

x_1 x_2 = -a

-a=-1

1=-a

a=-1, 1

The question is vague. It can be any odd multiple of 15.

01100011011011110111001001110010011001010110001101110100

i guess 15?

Use diagram below:

Answered here

How many ways are there to put 4 distinguishable balls into 3 indistinguishable boxes?

can the boxes be empty.  I will asume they can

Everything is indsitinguishable so...

4,0,0

3,1,0

2,2,0

2,1,1

that is it.    4 ways.

Thanks for answering BuilderBoi,

I'd like you to think about your presentation.

In my oppinion if you spread your ideas out they would be easier to comprehend.

'Blocks' always look intimidating to me.

Just a suggestion.

Consider the polynomial f(x) = 1 - 12x + 3x^2 - 4x^3 + 5x^4 and g(x) = 3 - 2x - 6x^3 + 19x^4. Find c such that the polynomial f(x) + cg(x) has degree 3.

\(f(x) = 1 - 12x + 3x^2 - 4x^3 + 5x^4 \)

\(g(x) = 3 - 2x - 6x^3 + 19x^4\)

Looking just at the x^4 term

   \(f(x) + cg(x)=..... 5x^4+c*19x^4\\ f(x) + cg(x)=..... (5+19c)x^4\)

if the x^4 term is to disappear then    5+19c=0     etc

Let's see

\(\sqrt{60x}*\sqrt{12x}*\sqrt{63x}*\sqrt{42x}\\ =\sqrt{60}*\sqrt{12}*\sqrt{63}*\sqrt{42}*\sqrt{x^4}\\ =2\sqrt{3*5}*2\sqrt{3}*3\sqrt{7}*\sqrt{7*6}*x^2\\ =2*2*3 *\sqrt{3*5*3*7*7*6}*x^2\\ =12 *3*7*\sqrt{5*6}*x^2\\ =12 *3*7*\sqrt{5*6}*x^2\\ =252\sqrt{30}\;x^2\)

There you go BilderBoi,    we are in full agreement   

Have you drawn the pic yet?

The angle that they have given you is not actually in the triangle.

Depending on which angle of the triangle that you use you will need to use either sin or cos to get the answer.

5x² - 11x + 4 = 0 => Let the roots be a, b => a + b = 11/5, ab = 4/5

a³ + b³ = (a + b)³ - 3(ab)(a + b) = (11/5)³ - 3(4/5)(11/5) = 1331/125 - 132/25 = (1331 - 660)/125 = 671/125.

x^2+2y-9=-y^2+18x+9+12x+22 => x² + y² - 30x + 2y - 40 = 0 => (x - 15)² + (y + 1)² = 266

Center ( 15, -1), radius r = \(sqrt{266}\)