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Thanks anon  :)

Be considerate of others!

You have asked this same questions twice and it has been answered twice.

This is a waste of our time and it means that someone else will not get their questions answered at all.

Be considerate of others!

Thanks Heureka

You have asked this same questions twice and it has been answered twice.

This is a waste of our time and it means that someone else will not get their questions answered at all.

Be considerate of others!

$$y=16x+44 \qquad and \qquad y= -x^2+8x+28$$

I can eliminate the ys right off

$$\\16x+44 =-x^2+8x+28\\
16x+44-16x-44 =-x^2+8x-16x+28-44\\
0=-x^2-8x-16\\
x^2+8x+16=0\\
(x+4)^2=0\\
x=-4\\\\
y=16*-4+44\\
y=-64+44\\
y=-20$$

x=-4    and y=-20

That looks impressive Chris  

Look in the sticky notes at the bottom right of your page - one is one fractions and the calc

That dialogue is really funny :))

With 1 head given for certain, you need exactly 8 more heads and 2 tails.

The probability of a given sequence of 8 heads and 2 tails is (1/2)⁸*(1/2)² = 1/1024.

There are nCr(10,8) = 45 ways of achieving such a sequence, so the overall probability is 45/1024 = 0.0439 to four decimal places.

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You can expand this to 10x*2x + 10x*3 → 20x² + 30x  

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But it's not here!

 its there in my textbook .buddy

What do you mean by the a*b formula?

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123456789

His clock says t1 when he leaves.  The time on his friends clock when he arrives is T, so his apparent journey time is  δt = T - t1.  He stays for a known time δT.

Assuming his apparent return journey time is also δt then his apparent round trip takes 2δt + δT or 2(T - t1) + δT. His clock now reads t2.

Assuming his clock runs at the right speed the interval Δt = t2-t1 is the true interval (though neither t1 nor t2 show the true time).

half the difference between the true interval and the apparent interval gives the required correction:

correction = (2(T-t1) + δT - Δt)/2 

All the values on the RHS are known, so he knows the size of the correction.  If it is positive the apparent interval is longer than the true interval, which means his clock is behind true time.  If it is negative the apparent interval is shorter than the true interval and his clock is ahead of true time.

So he knows by how much and in what direction to adjust his clock.

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You can go further:

7(x² + 5x + 6) = 7(x + 2)(x + 3)

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The E21 just means 10^21 so you have 639*10²¹/1.0659*10¹⁶ = 639*10^21-16/1.0659 = 639*10⁵/1.0659 = 6.39*10⁷/1.0659 ≈ 5.9949*10⁷

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Turn both sides upside down to get 117/72 = f/8

Multiply both sides by 8 to get 8*117/72 = f  or f = 13

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$$\frac{8000}{-875\times 2^{30}}$$ Can be evaluated using the calculator.

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There is no object above :(

If you need/want to display pics, you will need to be a member. :)

An excellent answer MG   

The pi button is next to the [2nd] button at the top

$${\mathtt{44.7}} = {\frac{{\mathtt{100}}}{{\mathtt{x}}}}$$

$${\mathtt{44.7}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{100}}$$

$${\mathtt{x}} = {\frac{{\mathtt{100}}}{{\mathtt{44.7}}}} = {\frac{{\mathtt{1\,000}}}{{\mathtt{447}}}}$$

Thanks Chris and TheMathsStudent

I thought there was some definative way to do this.   

$$\\7Xmod26=1\\\\
\frac{7x}{26}=N+\frac{1}{26} \qquad $Where X and N are both integers$\\\\
7X=26N+1$$

Multiples of 26 are   26,52,78, 104,130, 156, 182,

(Multiples of 26) +1  are   27,53,79, 105,   105 = 7*15

So X=15 works this is one solution

7*15=26(N)+1        N=4

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$$\left({\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{15}}\right) {mod} \left({\mathtt{26}}\right) = {\mathtt{1}}$$

That is good   X=15 is the smallest value of X

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Now I want to find a general solution   

If (7*15) mod 26=1

Then I think

(7*15+26G) mod 26 must also =1

Thinking,  thinking.      :/