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You can do at least some of this yourself!

Come back with your attempt

I believe the center of the circle has to be equidistant to (12, 0) and (0, 8). 

So the raduis could be any length greater than 5sqrt(2). 

But I'm probably missing something. 

Sorry I couldn't help more. :))

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The raduis of the circle is 11, so the total area is 121pi (pi*r^2)

60 degrees is 1/6 of the circle. (60/360)

So the answer is 1/6 of 121pi. 

I hope this helped. :))

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Hmmm, I don't think I understand why. 

Could you explain?

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This is an equation of a circle, where the center is at the origin and radius. 

Since all the x points are mirrored at the y axis, I believe it should sum up to 0. (for example if x = 3, x can also be -3, and 3-3 = 0). 

I hope this helped. :)))

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The answer actually is divide by 1, i guessed it and got it right.

y = 6x + b

We're solving for b so we can plug things in. 

19 = 6(2) + b

19 = 12 + b

b = 7

So our equation is y = 6x + 7

I hope this helped. :)))

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Why not?

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I dont think it is right.

There are 9 options for savory, and 5 for sweet. 

So we just need to multiply together. 

9*5 = 45. 

I hope this helped. :)))

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Hello!

2x + 5 = 3x - 4 (subtract 2x)

5 = x - 4 (add 4)

9 = x

Not sure what the last step is, but that's how I would solve it. 

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(x)^2 + (x+y)^2 = (x + 2y)^2 (this is the pythagreon theorum)

x^2 + x^2 + y^2 + 2xy = x^2 + 4y^2 + 4xy

2x^2 + y^2 + 2xy = x^2 + 4y^2 + 4xy

x^2 = 3y^2 + 2xy

So we're looking for xy, and we could solve for it, but let's just say x = 1 for simplicity.

1^2 = 3*y^2 + 2(1)y

1 = 3y^2 + 2y. 

y = 1/3. 

So y/x = 1/3/1 = 1/3. 

I hope this helped. :)))

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\(x^{log_y{x}}=2  \qquad  \qquad \qquad y^{log_x{y}}=2^4\\ x^{\frac{log x}{log y}}=2  \qquad  \qquad \qquad y^{\frac{log y}{log x}}=2^4\\ log\;{x^{\frac{log x}{log y}}}=log\;{2}  \qquad  \quad \log{y^{\frac{log y}{log x}}}=\log{2^4}\\ {\frac{log x}{log y}}*log\;x=log\;{2}  \qquad  \quad {\frac{log y}{log x}}*\log\;y=4\log\;2\\ \frac{({log x})^2}{log y}=log\;{2}  \qquad  \quad \qquad \frac{(log y)^2}{4log x}=\log\;2\\~\\ so\\ 4{{(log x})^3}=(log y)^3\\ \sqrt[3]{4}=\frac{logy}{logx}\\ log_xy = \sqrt[3]{4}\)

Oh I have to solve for x...

\(\frac{logx}{log y}=\frac{1}{\sqrt[3]{4}}\\ log_y x=4^{(-1/3)}\\ y^{log_y x}=y^{4^{(-1/3)}}\\ x=y^{4^{(-1/3)}}\\ \)

First house has a chance of 1/5. (1 mail is correct, you have 5 mails.)

Second house has a chance of 1/4. (1 mail is correct, you have 4 mails, only 4 since you have you already delivered the first house.)

Third house has a chance of 1/3. 

Fourth house has a chance of 1/2. 

Fifth house has a chance of 1/1. 

What's 1/5*1/4*1/3*1/2*1/1?

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3 sides in a triangle. 

2 of the sides are 15 and 20. 

The whole 3 sides is 50.

What's 50 - 15 - 20?

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Hint: Cross multiply, expand, then solve the quadratic

Thanks for your input grs75.

If you did it yourself then you have more skill and knowledge than you are giving yourself credit for.

Line YW = \(\sqrt{26^2 - 24^2} = 10\) 

This means XY and VY are also 10 units. 

SX = \(\sqrt{29^2 - 10^2} = \sqrt{741} \)

This means VS is also \(\sqrt{741}\)

VP = 24

Let the number of kg of nickel alloy used=a
(33*0.35 + 0.21a)=(33 + a)*0.27, solve for a
11.55 + 0.21a =8.91 +0.27a
11.55 - 8.91 =0.27a - 0.21a
2.64 =0.06a
a =2.64 / 0.06
a =44 kg of nickel alloy were used.
33 kg of 0.35 pure nickel + 44 kg of 0.21 pure nickel alloy = 77 kg of 0.27 pure nickel alloy.

Nice Job @idyllic! I would use a different strategy, first of all do you know addition then this is another tedious strategy for multiplying with large numbers.

The strategy is skip-counting, we add 7 7 times because 7 * 7, and we get

7+7+7+7+7+7+7+7 = 49.

Please tell me if you have any trouble with understanding this.

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Thank you very much! This one stumped me for a long time :)

Here's what I came up with a while  back :

164x + 37y  =  1

164 = 4*37 + 16

37 =  2*16 + 5

16 = 3*5 +  1

Working backwards, we have

1 = 16 - 3(5)

1 = 16 - 3[ 37 - 2(16) ]

1 = -3(37) + 7(16)

1 = -3(37) + 7 [ 164 - 4*37 ]

1 =  -3(37) + 7(164) - 28 (37)

1 = 164(7) + 37(-31)

(x, y) = (7, -31)

Thx , Idyllic  !!!!