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  A

      4            5

     B        D          C

                3

Since AD is a  bisector.....Let BD = x  and DC  =3 - x   and  we have that

BD/ AB  = DC / AC

x / 4 = (3 - x) / 5

5x = 4 (3 - x)

5x  = 12 - 4x

9x= 12

x =  4/3  = BD

[ ABD ] = (1/2) (4/3)(4)  =  16/6 = 8/3

[ABC ]  = (1/2)(4)(3)  = 6

[ADC ]   =  [ ABC ] -  [ ABD ]  = 6 - 8/3  =   10/3 =  3 + 1/3  =   3 (to the nearest integer )

$a+ar+ar^2+...=\frac{a}{1-r}$ as long as r<1

$a^3+a^3(r^3)+a^3(r^3)^2+...=\frac{a^3}{1-r^3}$  as long as r^3<1

So:  

$\frac{a}{1-r}=4$

and

$\frac{a^3}{1-r^3}=10$

Can you take it from here?

(x+8)^2=86

We try to pick out all the combinations of terms, one from each polynomial, such that their product is some number times $x^2$.

First, if we pick 2x^3 from the first polynomial, it is impossible to get x^2 by multiplying it to another term from the other polynomial, we do not care about this case in this scenario.

Then, we consider the second term x^2 from the first polynomial. To get some multiple of x^2, we can only choose the constant term from the second polynomial, which will give us $+15x^2$ if we multiply the terms.

Then, for -3x, we choose -11x from the second polynomial, so when we multiply them, we get $+33x^2$

For +5, we choose tx^2 from the second polynomial, so when we multiply them we get $+5tx^2$.

Therefore, the term in x^2 is $15x^2 + 33x^2 + 5tx^2$, which is $(5t + 48)x^2$. Since we are given that the product has no x^2 term, the coefficient must be 0, which gives $t= -\dfrac{48}5$.

$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10}$ is basically:

1/3(1 - 1/4 + 1/4 -1/7 +1/7 - 1/10)

which is 1/3 $\times$ 9/10 = 3/10

If I am not mistaken, this is the diagram to corresponding to your question. We can then find a because of the inradius formula (a+b-c)/2 so a=3.

the sides then form a pythagorean triple 8,15,17 and 8+15+17 = 40, the answer.

Considering the sum of interior angles of the triangle we have $\angle B = 34^\circ$. Then we have, by sine rule,

$\dfrac{5}{\sin 34^\circ} = \dfrac{\overline{BC}}{\sin 108^\circ}$

From here, we have $\overline{BC} \approx8.50$.

Simplifying, the quadratic equation is $x^2 + 20x - 13 = 0$. Therefore, a + b = -20 and ab = -13.

Now, 

$\quad (a^2 + a + 1) + (b^2 + b + 1)\\ = (a^2 + b^2) + (a + b) + 2\\ = (a + b)^2 - 2ab + (a + b) + 2\\ = 400 + 26 - 20 + 2\\ = 408$

Also,

$\quad (a^2 + a + 1)(b^2 + b + 1)\\ = \dfrac{(a^3 -1)(b^3 - 1)}{(a - 1)(b - 1)}\\ = \dfrac{(ab)^3 - (a^3 + b^3) + 1}{ab - (a + b) + 1}\\ = \dfrac{(ab)^3 - (a + b)^3 + 3ab(a+b) + 1}{ab - (a + b) + 1}\\ = 823$

Hence, the required equation is $x^2 - 408x + 823 = 0$.

Simplifying, the equation you have is $x^2 - 23x - 16 = 0$.

From Vieta's formulas, we have $\begin{cases} a + b = 23\\ ab = -16 \end{cases}$.

Now, note that $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2 b^2 = ((a+b)^2 -2ab)^2 - 2(ab)^2$.

Can you continue from here?

The value of angle does this replacement ramp make with the horizontal is,

⇒ θ = 14.1°

A 30-foot ramp is used to move product to a loading dock.

The ramp makes a 24° angle with the horizontal.

And, The ramp is to be replaced with a ramp that is 50 feet long.

We have;

⇒ sin θ = 30 × sin 24° / 50

⇒ θ = sin ⁻¹ (30 × sin 24° / 50)

⇒ θ = 14.1°

difference of sines:

sin(pi/6)-sin(pi/4)=

sin(pi/6)cos(pi/4)-sin(pi/4)cos(pi/6)

and try finishing it off

the answer is (x+8)2=86

the answer is 

−1.25x^2+7.5x=0

11.83

we can notice that 8, 12, and 15 are pythagorean triples, so the points A,B, and C form a right triangle ABC. Since ABC is inscribed in the circle, we know that its hypoteneuse, 15, is the diameter of the circle.

Therefore, $\boxed{15/2}$ is the radius of the circle.