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distance b/w Danny and Hot Racer = 40 meters

⇒ 40  = \( {5 \over 6}\frac{}{}\)* track's length

⇒ track's length = 40 * \( {6 \over 5}\frac{}{}\)

⇒ track's length = 8*6 = 48 meter

If it has to be positive, then only 1 pair? (223, 3)?

i think thats incorrect

The equation of the line is y = 4x - 1, so the inequality is y <= 4x - 1.

t = track length

5/6 * t =   40      multiply both sides by 6/5

t = 40 * 6/5 = 48 meters

x + 3y = -5               re arrange to y = mx+b from:

y = -1/3 x - 5/3         slope = -1/3     ( parallel slope is the same)

y = -1/3 x -  b           sub in the given point to calculate 'b'

2 = -1/3 (-7) + b       shows b = - 1/3

y = -1/3 x - 1/3         multiply by 3 to get rid of fraction

3y = 3x - 3

3x -3y = 3                 then   B - A =  -3 - 3 = - 6

look for a pattern

106 divisible by 3    no

116   no

126   yes

136   no

146   no

156   yes

166   no

176    no

186     yes                 looks like every third one is divisible by 3            = 1/3

The minutes hand would be on the 6, and the hours hand would be halfway between 2 and 3, because 30 is half of 60.

A circle is 360 degrees, so each gap between the numbers is 30 degrees each.

$30 \cdot 3.5 = 105$, so the answer is 105 degrees

Putting the equations into slope-intercept form,

\(y = -\frac{Ax}{B} + \frac{3}{B}\)

\(y = -\frac{x}{3} - \frac{5}{3}\)

Both lines are parallel, so \(-\frac{x}{3} = -\frac{Ax}{B}\). Simplify to find that \(\frac{A}{B} = \frac{1}{3}\), meaning that the slope is \(-\frac{1}{3}\).

\(y = -\frac{x}{3} + \frac{3}{B}\)

Plug in (-7, 2).

\(2 = -\frac{-7}{3} + \frac{3}{B}\)

\(6B = 7B + 9\)

\(B = -9\)

Because A and B are in ratio \(\frac{A}{B} = \frac{1}{3}\), B = -9 and A = -3. So B - A = -9 - (-3) = -6

i'm afraid thats wrong

From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, B - A = -19/3 - 3 = -28/3.

This is  an infinite  geometric  sum   with  r =   2/3

S =  (1/3)   / ( 1 - 2/3)  =   (1/3) / ( 1/3)  =    1

Well we can simplify both of them starting with √50.

\(\sqrt{50}\)

\(=\sqrt{2\cdot \:5^2}\)

\(=\sqrt{2}\sqrt{5^2}\)

\(=5\sqrt{2}\)

\(\sqrt{18}\)

\(=\sqrt{2\cdot \:3^2}\)

\(=\sqrt{2}\sqrt{3^2}\)

\(=3\sqrt{2}\)

So adding \(3\sqrt{2}+5\sqrt{2},\) you get \(8\sqrt{2}\)

Rewrite it with simplified radicals,

\(5\sqrt{2} + 3\sqrt{2}\)

They both have a radical of 2, so we can combine them to get \(8\sqrt{2}\)

To find the bases, we can plug in our known values with x being the smaller base,

\(\frac{2x+9}{2}\cdot 8 = 60\)

\(\frac{2x+9}{2}=7.5\)

\(2x+9 = 15\)

\(x = 3\)

The smaller base is 3, and the larger is 12. The median of a trapezoid is just the average of the bases. So (3+12)/2 = 7.5 which is our answer.

$n \leq 1000$

$n \equiv 7 \pmod{9}$

$n \equiv 5 \pmod{10}$

$n \equiv 9 \pmod{11}$

By the chinese remainder theorem, you have 

$n = 990k + 295$

$n \leq 1000$

$n = \boxed{295}$

Only  one little error......

sin x  =  3/5     

cos  x =  4/5

a = (4/5 , 3/5)   this is OK

Rememnber  that   pi/2 < b <  pi....so....the  cos is negative  in ths quadrant

sin x  =  4/5

cos x=  -3/5

b= (-3/5, 4/5)

c1, c2 ,d1 , d2   =       4/5 , 3/5 , -3/5 , 4/5

Your answer seems correct.....maybe  the way you entered it  ???

You could have also noted that the units digit is simply \(\begin{align*} 3^3 &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= \boxed{7} \end{align*}\)

Too bad I can't use \equiv on begin{align*} enviorments.

Let $f(x)$ equal units digit of $x.$

$f(3^1) = 3$

$f(3^2) = 9 $

\(\begin{align*} f(3^3) &= 27 \pmod{10} \\ &= 7 \pmod{10} \\ &= 7 \end{align*}\)

\(\begin{align*} f(3^4) &= 81 \pmod{10} \\ &= 1 \pmod{10} \\ &= 1 \end{align*}\)

\(\begin{align*} f(3^5) &= 243 \pmod{10} \\ &= 3 \pmod{10} \\ &= 3 \end{align*}\)

Thus, given $3^n,$

$n \equiv 0 \pmod 4, f(n) = 1$

$n \equiv 1 \pmod 4, f(n) = 3$

$n \equiv 2 \pmod 4, f(n) = 9$

$n \equiv 3 \pmod 4, f(n) = 7$

$123 \equiv 3 \pmod 4, $ so $f(n) = \boxed{7}$

thanks!

I don't know if this is by all of these, but I'm assuming the a,b,c,d are not constraints for one question but multiple questions.

(a) $3 + 5 + 2 + 9 + a + 2 = 3n \Rightarrow 21 + a = 3n \Rightarrow a=0; a=3; a=6; a=9$

(b) $a+2 = 4n \Rightarrow a = 2; a = 6$

(c) All values of (b) that are also in (a), or vice versa $\Rightarrow a = 6$

(d) All values of (c) that are also in (a), or vice versa $\Rightarrow a = 6$

oh shoot lol

ummm...^ means exponent...

We can find a pattern in the exponents. Because we are only looking for units, let's find a pattern in 3s.

3^1, 3^2, 3^3, 3^4, 3^5

3, 9, 7, 1, 3

So every 4th multiple will have a units digit of 1. 120 is divisible by 4, meaning 123^120 has a units digit of 1, so 123^123 will have a units digit of 7.

THANKS!