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What is the residue of \(9^{2010}\), modulo \(17\) ?

\(\begin{array}{|rcll|} \hline && \mathbf{9^{2010} \pmod {17}} \\ &\equiv & \left(3^2 \right)^{2010} \pmod {17} \\ &\equiv & 3^{2\cdot 2010} \pmod {17} \\ &\equiv & 3^{4020} \pmod {17} \\ & & \boxed{ a^{\varphi(n)} \equiv 1 \pmod {n} \text{, if gcd(a,n)=1 } \\ 3^{\varphi(17)} \equiv 1 \pmod {17} \text{, gcd(3,17)=1 } \quad \varphi(17) = 16 \\ 3^{16} \equiv 1 \pmod {17} } \\ &\equiv & 3^{16\cdot 251+4} \pmod {17} \\ &\equiv & \left(3^{16} \right)^{251}3^4 \pmod {17} \\ &\equiv & \left(1 \right)^{251}3^4 \pmod {17} \\ &\equiv & 3^4 \pmod {17} \\ &\equiv & 81 \pmod {17} \\ &\equiv & \mathbf{13 \pmod {17}} \\ \hline \end{array}\)

Two circles of radius \(1\) are centered at \((4,0)\) and \((-4, 0)\).
How many circles are tangent to both of the given circles and also pass through the point \((0, 5)\)?

4 circles are tangent to both of the given circles and also pass through the point \((0, 5)\):\(\begin{array}{|l|lcll|} \hline 1 & x^2+\left(y-\dfrac{5}{3}\right)^2=\left(\dfrac{10}{3}\right)^2 \\ \hline 2 & \left(x+1.0328\right)^2+\left(y-1\right)^2\ =\ \left(4.13118\right)^2 \\ \hline 3 & \left(x-1.0328\right)^2+\left(y-1\right)^2=\left(4.13118\right)^2 \\ \hline 4 & x^2+y^2=5^2 \\ \hline \end{array} \)

Take a look:

https://web2.0calc.com/questions/help_53566

Take a look at:

https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_21

I think it's the problem you're looking for.

oh thank you i get it now

\(\sqrt{a}-\sqrt{b}=20\\ a = (20+\sqrt{b})^2\\ a-5b = (20+\sqrt{b})^2-5b\\ \dfrac{d}{db} (20+\sqrt{b})^2-5b = \\ 2(20+\sqrt{b})\cdot \dfrac{1}{2\sqrt{b}} - 5\\ \text{and we set this equal to 0}\)

\(\dfrac{20}{\sqrt{b}}+1 = 5\\ \sqrt{b}=5\\ b=25\\ \sqrt{a}-5=20\\ a=625\)

\(\text{Max value of $a-5b=625-5(25)=500$}\)

How many integers n satisfy \(\large{(n+3)(n-7) \leq 0}\)?

Tip:

It's Traviso

Don't forget the A!

Ok so I know that:
\(\cos (x)=\sin\left(x+ \dfrac{\pi}{2}\right)\) (Radians)
what is:
\(\sin(x)=\cos(\ldots)\)?

\(\begin{array}{|rcll|} \hline \cos (x) &=& \sin\left(x+ \dfrac{\pi}{2}\right) \\ && \boxed{ x+ \dfrac{\pi}{2} = x' \quad | \quad - \dfrac{\pi}{2} \\ x = x'- \dfrac{\pi}{2} } \\ \cos (x'- \dfrac{\pi}{2}) &=& \sin\left(x'\right) \\ \sin(x') &=& \cos\left(x'- \dfrac{\pi}{2}\right) \\ \hline \end{array} \)

\(\mathbf{\sin(x)=\cos\left(x- \dfrac{\pi}{2}\right)} \)

Please show us some working or thought processes trvisio.

How many integers n satisfy \(\large{(n+3)(n-7) \leq 0}\)?

We create a sign table:

\(\begin{array}{|l|c|c|c|c|} \hline \text{Interval or position} : & (-\infty,-3) & -3 & (-3,7) & 7 & (7,\infty) \\ \hline \text{sign of } (n+3): & - & 0 & + & + & + \\ \hline \text{sign of } (n-7): & - & - & - & 0 & + \\ \hline \text{sign of }(n+3)(n-7): & + & \color{red}0 & \color{red}- & \color{red}0 & + \\ \hline \end{array}\)

We can read the result: in the interval [-3,7] the left side of the inequality is negative or zero, and thus the inequality is true there.

\(-3\leq n \leq 7 \qquad n=\{-3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7 \} \quad (11 \text{ integers, assumption zero is an integer }) \)

There are a couple of other ways....   this is a bowl shaped parabola (because the leading coefficient is positive)

- b / 2a

= 9/2

or take the derivative and = 0

2t-9 = 0

2t=9

t=9/2

Deleted ....never mind

Try drawing the two triangles together.  They have 2 sides that are the same length.

Both have the same base.   WX=WZ

Both have an adjoining side that is the same length. Shown by the arc.

If i make the angle between them bigger, what happens to the third side?

A conditional statement is a statement that can be expressed as "If _____, then _____."

The statement  "I will eat anything that is dipped in chocolate,"  can be expressed as  "If something is dipped in chocolate, then I will eat it."

yes and she didnt toss me in front of a truck. I acciendently walked in front of it...

long story.

And I havent done that ever since.

Oh ok Thank you Hecticar!

Three questions in a row, all very similar, who do you think your are kidding?

This was the first one, I will grant you that.   

it means on average you win less than the 50c you paid to play.

so it is A

To find the minimum value, you can complete the square.

Using the formula \(a(t+d)^2+e\) where \(d=\frac{-b}{2a}\) and \(e=c-\frac{b^2}{4a}\), the completed square would be \((t-\frac{9}{2})^2-45\). Since squares cannot be negative, we see that the minimum value would be where \((t-\frac{9}{2})^2\) is 0. Thus \(t=\frac{9}{2}\)

This diagram has more information than I showed how to find. If you're unsure how to find, for example,  AD = 30, and want further explanation, just ask.

m∠EDC  =  m∠ABC   because corresponding angles are congruent.

m∠ECD  =  m∠ACB   because they are the same angle

So by the AA similarity theorem,  △EDC ~ △ABC

When we draw a height to the base of an isosceles triangle, we split the base exactly in half.

So the scale factor from △ABC  to  △EDC  is  1/2.  And so...

area of △EDC  =  ( 1/2 )( 1/2 )( area of △ABC)

area of ABDE   =   area of △ABC - area of △EDC

area of ABDE   =   180 - ( 1/2 )( 1/2 )( 180 )

area of ABDE   =   180 - 45

area of ABDE   =   135

The areas are in square units.

Yeah I think so