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Just divide:

13,180 / 66,104 = 0.20 or 0.20 x 100 = 20% of the population.

Solve for r:
(r + 3)/(r - 2) = (r - 1)/(r + 1)

Cross multiply:
(r + 1) (r + 3) = (r - 2) (r - 1)

Expand out terms of the left hand side:
r^2 + 4 r + 3 = (r - 2) (r - 1)

Expand out terms of the right hand side:
r^2 + 4 r + 3 = r^2 - 3 r + 2

Subtract r^2 - 3 r + 2 from both sides:
7 r + 1 = 0

Subtract 1 from both sides:
7 r = -1

Divide both sides by 7:

 r = -1/7

10pm till 6am = 8 hours that the lamp is turned on.

8 x 9 watts =72 watts consumed in 1 day.

72 / 1000 =0.072 Kilo-watt/hours consumed in 1 day.

0.072 x $0.20 =$0.0144 - cost of electricity for 1 day.

$0.0144 x 365 days =$5.26 cost of electricity for the Smith family in 1 year.

!Gracias!

We can solve this by spliting the trapazoid into one rectangle, and two triangles of exact same area. 

First lets start of witht he triangles, we know that the area of a triangle is\(\frac{1}{2}* Base * height = area\) 

We can see that both triangles have a base of 4 inches, and a height of 14 inches  

So we can plugh that in our equation. \(\frac{1}{2}*\frac{4}{1} *\frac{14}{1} = \frac{56}{2} = 28\)

and we can multiply that by 2 which is 56 

Finally all we have to check is the rectangle it's 14 inches long we can see and it 10 inches wide and we know the formula for an area for a rectangle is \(Lengh * With = area\) 

so \(14 * 10 = 140\) So we have 140 now and all we need to do is add it to the 56 we originally had. 

Which leaves us with 196 so you were correct : )

So I was right yeah 

(CELEBRATION NIGHT COME ON ITS A CELEBRATION.... 

you can reaarange the trapezoid to form a rectangle by flipping one of the triangles and placing next to the other. The measurements of the rectangle will be 14 inches by 14 inches, so 196 square inches 

The triangle on the right is 

14*4=56 

and 56/2=28  

so each Triangle is 28sqin 

The Square is 10*14= 140sqin 

140+28+28=196sq in 

It says the answer is incorrect, do you know where you went wrong?

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a)2a^4-7a^3+8a^2-5a+3

b)2y^4+5y^3-7y^2+23y-30

c)x^4-2x^3+3x^2-2x+1

d)x^4+4x^3-2x^2-12x+9

Let me know if you'd like working!

\(p(x) = x^2 + c_1 x + c_0 \Rightarrow r_1 + r_2 = -c_1\\ f(x) = x^2 - a x + f_0\\ g(x) = x^2 - b x+ g_0\\ h(x) = x^2 - c x + h_0\\ y(x) = f(x) + g(x) + h(x) = 3x^2 -(a+b+c)x + (f_0 +g_0+h_0)\)

\(y(x) = x^2 - \dfrac 1 3(a+b+c)x + \dfrac 1 3 (f_0 + g_0 + h_0)\\ \text{This is the same form as $p(x)$ listed above with $c_1=-\dfrac 1 3(a+b+c)\\$and thus the sum of the roots is $\dfrac 1 3 (a+b+c)$}\)

OHHHH, that makes so much more sense, thank you so much hectictar!! :)

6C3=20

\(d\ =\ \sqrt{(1-5)^2+(\sqrt2-\sqrt8)^2} \)          But   \(\sqrt2-\sqrt8\ \neq\ -2\)    !!

\(d\ =\ \sqrt{(1-5)^2+(\sqrt2-2\sqrt2)^2}\)          because   \(\sqrt8\ =\ 2\sqrt{2}\)

\(d\ =\ \sqrt{(1-5)^2+(-\sqrt2\,)^2}\)          because   x - 2x  =  -x    so     \(\sqrt2-2\sqrt2\ =\ -\sqrt2\)

\(d\ =\ \sqrt{(-4)^2+(-\sqrt2\,)^2}\)          because  1 - 5  =  -4

\(d\ =\ \sqrt{16+2}\)          because   \((-4)^2=16\)     and     \((-\sqrt2)^2=2\)

\(d\ =\ \sqrt{18}\)

\(d\ =\ 3 \sqrt{2}\)_

\(\left(\dfrac 1 a + \dfrac 1 b\right)\left(\dfrac 1 b -\dfrac 1 a\right) = \dfrac{1}{b^2}-\dfrac{1}{a^2}\)

\(\text{to minimize this we want the largest possible magnitude of $b\\$and the smallest possible magnitude of $a$}\\ a = -1,~b=3\\ \dfrac{1}{3^2}-\dfrac{1}{(-1)^2} = \dfrac 1 9 - 1= -\dfrac 8 9\)

how many segments will he draw before returning to his starting point at A 

Nobody has answered this yet, so I'm going to take a run at it. 

Never mind the inside circle; we don't need it.  Just draw each successive chord at a 75^o angle.  I went ahead and starting drawing them, and the end of the tenth chord landed at A again.  Stuart will draw a total of 10 lines (i.e., an additional 7 to the 3 that were already there).

.

The total number of students: 10 C6=210.

Boys: 6C3 and Girls: 4C3=4, 20*4=80.'

Thus, the answer is 80/210=8/21. 

-tertre

\(\text{If $p(x)$ isn't monic, then normalize it so it is. $\\$This won't affect the roots or the sum of the coefficients being 0}\\ p(x) = x^2 + b x -(1+b)\\ p(x) = (x-1)(x+1+b) \text{ with roots}\\ x=1,~x=-(1+b) \text{, both of which are real}\)

I'm not quite sure about how to solve a problem like this, but I'll mess around with it and see what I get.

Lets try to factor it by grouping. To factor nicely the 11x should split up with one of the parts a multiple of 5.

5x^2+5x+6x+c

5x(x+1)+6(x+c/6)

If this is going to be factored, the second part needs to match the first part. For this to happen c/6=1 so c=6.

(5x+6)(x+1)

The solitutions of this are x=-1 and x=-6/5. These are both rational.

There is one more way 11x can be split up with a multiple of 5.

5x^2+10x+x+c

5x(x+2)+1(x+c/1)

For this one to factor, c/1=2 so c=2.

(5x+1)(x+2)

The solitutions to this one are x=-2 and x=-1/5 and they are also rational.

So when c=6 and c=2 the quadratic has rational solitutions and the answer you are looking for is \(2\cdot6=\boxed{12}\)

ratio of your first triangle is 7-24-25 (it's a pythagorean triple, you could also find 25 using pythagoreon thereom) 

ratio of bigger tirangle is x:x:100. To get from 25 to 100, x4. 7 x 4 = 28, 24 x 4 = 96

The ratio of your bigger triangle is 28 : 96 : 100

The shorter leg value of the second triangle is 28

By the pathogorean theorum, the hypotonuse of the first triangle is \(\sqrt{7^2+24^2} = 25\)

Because the two triangles are similar, the rario between the side lengths will be the same on both triangles.

\(\frac{100}{short leg}=\frac{25}{7}\) 

cross multiply

700=25*short leg

divide by 25

short leg of the second triangle=\(\boxed{28}\)

A) 15 / sin 43 degrees. = 13 / sinB       rearrange as

sin B = 13 sin 43 / 15    =   (13/15)sin43

arcsin [ (13/15) sin 43 ]  = B  ≈  36.2°

C) sin14 degrees / 3.1 = sin27 degrees / b        rearrange as

b  = (3.1)sin 27/ sin 14  ≈  5.8