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okay got it!

The image uploader wasn't working for me so I just posted pictures of the problem here.

Looks to me as though none of the specified possible solutions apply!!

No zeros in sight!

You are right, So I did.     

Thanks for picking that up.  :)

Obtuse scalene triangle.
Sides: a = 13 b = 37 c = 40

Area: T = 240
Perimeter: p = 90
Semiperimeter: s = 45
Calculate the heights of the triangle from its area.
There are many ways to find the height of the triangle. The easiest way is from the area and base length. The area of a triangle is half of the product of the length of the base and the height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.
Height =[2 x Area] / Longest side
Height =[2 x 240] / 40 =12

Hi Melody: I think you made a typo in the SUM of the sequence. It is 100^100 not 10^10.

100/2 * [2F + 2*(100 - 1)]=100^100, solve for F

F =195 x 9 + 901 = 198 digits long =~ 1 x 10^198

One root of \(x^2+12x+k=0\) is twice the other root.

Find \(k\).

Let the roots are \(x_1\), \(x_2\).

\(\begin{array}{|rcll|} \hline x^2+12x+k=0 &=& (x-x_1)(x-x_2)= 0 \quad | \quad x_2 = 2x_1 \\\\ &=& (x-x_1)\Big(x-(2x_1)\Big) \\ &=& (x-x_1)(x-2x_1) \\ &=& x^2-2x_1x-x_1x+2x^2_1 \\ x^2+12x+k=0 &=& x^2-3x_1x+2x^2_1 \\ && \text{compare} \\ && \boxed{12=-3x_1\ \text{ or }\ x_1=-4 \\ k= 2x^2_1\ \text{ or }\ k=2*(-4)^2=32 } \\ \hline \end{array} \)

\(\boxed{k=32}\)

"In base 8, the four-digit numeral BBCC is the square of the two-digit numeral AA.  Find the ordered tirple of digits (A,B,C)."

x+7=0

    x=-7

(-7,0)

Which statement best reflects the solution(s) of the equation?

\(\sqrt{2x-4}-x+6=0 \)

A. There are two solutions: x = 4 and x =10.

B. There is only one solution: x = 10. The solution x = 4 is an extraneous solution.

C. There is only one solution: x = 4. The solution x = 10 is an extraneous solution.

D. There is only one solution: x = 10. The solution x = 0 is an extraneous solution.

Welche Aussage spiegelt die Lösung (en) der Gleichung am besten wider?

\(\sqrt{2x-4}-x+6=0 \)

A. Es gibt zwei Lösungen: x = 4 und x = 10.
B. Es gibt nur eine Lösung: x = 10. Die Lösung x = 4 ist eine Fremdlösung.
C. Es gibt nur eine Lösung: x = 4. Die Lösung x = 10 ist eine Fremdlösung.
D. Es gibt nur eine Lösung: x = 10. Die Lösung x = 0 ist eine Fremdlösung.

Hello jiminblossoms!
 
\(\color{BrickRed}\sqrt{2x-4}-x+6=0\\ \sqrt{2x-4}=x-6\\ (2x-4)=x^2-12x+36\)

\(x^2-14x+40=0\\ x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=7\pm\sqrt{49-40}\)

\(x_1=10\\ x_2=4\)

A. There are two solutions: x = 4 and x =10.

A. Es gibt zwei Lösungen: x = 4 und x = 10.

  !

The sum of two of the three roots of \(x^3 + ax^2 + bx + c = 0 \) is \(0\). 

Write an equation expressing c explicitly in terms of \(a\) and \(b\).

Let the roots are \(x_1,\) \(x_2\) and \(x_3\).

\(\begin{array}{|rcll|} \hline x^3 + ax^2 + bx + c &=& (x-x_1)(x-x_2)(x-x_3) = 0 \quad | \quad x_1+x_2 = 0\ \text{ or }\ x_2 = -x_1 \\\\ &=& (x-x_1)\Big(x-(-x_1)\Big)(x-x_3) \\ &=& (x-x_1)(x+x_1)(x-x_3) \\ &=& (x^2-x^2_1)(x-x_3) \\ x^3 + ax^2 + bx + c &=& x^3-x_3x^2-x^2_1x+x^2_1x_3 \\ && \text{compare} \\ && \boxed{a=-x_3 \text{ or } x_3=-a \\ b=-x^2_1 \text{ or } x^2_1=-b \\ c= x^2_1x_3 \text{ or } c=(-b)(-a)=ab } \\ \hline \end{array}\)

\(\boxed{c=ab} \)

I can do it a third way, although to be honest I do not know what Vieta's formula is

Sum of the roots in any polynomial is  = -b/a

In this case that is  -12/1 = -12

let one root be T the other is twice the size so   

T+2T=-12

3T=-12

T= -4

So the roots are  -4 and -8  so k is 32

If I understand the problem properly then I think that there are 4 possibilities for each combination of 4 colours/

There are 6C4 = 15 ways to choose those 4 colours

So I get 4*15 = 60 

If

\(a + b + c = 0 \)

and

\(a^3 + b^3 + c^3 = 216\),

find the value of \(abc\).

Formula:  \(\boxed{a^3+b^3+c^3 = 3abc+(a+b+c)(a^2+b^2+c^3-ab-bc-ca)}\)

\(\begin{array}{|rcll|} \hline \underbrace{a^3+b^3+c^3}_{=216} &=& 3abc+\underbrace{(a+b+c)}_{=0}(a^2+b^2+c^3-ab-bc-ca) \\\\ 216 &=& 3abc+0 \\ 3abc &=& 216 \quad | \quad : 3 \\ \mathbf{abc} &=&\mathbf{72} \\ \hline \end{array}\)

There are  9*10*10*10 = 9000  4 digit numbers

There are  9*9*8*7 = 4536 that do not have repeating digits.

SO there are 9000-4536 = 4464 that do

x = 4464/9000*100 = 49.6

Gore, I know you are very capable but on this occasion you did not read the question properly.

\(S_n=\frac{n}{2}(2a+(n-1)d)\\~\\ 10^{10}=\frac{100}{2}(2a+(100-1)*2)\\~\\ 10^{10}=\frac{100}{2}(2a+99*2)\\~\\ 10^{10}=100(a+99)\\~\\ 10^{8}=a+99\\~\\ a=10^{8}-99\\~\\\)

Check

\(S_n=\frac{n}{2}(2a+(n-1)d)\\~\\ S_{100}=\frac{100}{2}(2(10^8-99)+99*2)\\~\\ S_{100}=100(10^8-99+99)\\~\\ S_{100}=100(10^8)\\~\\ S_{100}=100^{10} \)

Of course this is not a great check because I have used exactly the same formula as I did when i worked it out.

S_n=\frac{n}{2}(2a+(n-1)d)\\~\\

10^{10}=\frac{100}{2}(2a+(100-1)*2)\\~\\

10^{10}=\frac{100}{2}(2a+99*2)\\~\\

10^{10}=100(a+99)\\~\\

10^{8}=a+99\\~\\

a=10^{8}-99\\~\\

Find the length of the line segments whose endpoints have polar coordinate (7, 40 degres) and (15, 100 degrees).

\(\text{Let $P_1 =\dbinom{x_1}{y_1}=\dbinom{7\cos(40^\circ)}{7\sin(40^\circ) } $} \\ \text{Let $P_2 =\dbinom{x_2}{y_2}=\dbinom{15\cos(100^\circ)}{15\sin(100^\circ)} $} \)

\(\begin{array}{|rcll|} \hline \mathbf{ \overline{P_1P_2} } &=& \mathbf{ \sqrt{ \left( x_2-x_1\right)^2 + \left(y_2-y_1 \right)^2 } } \\\\ &=& \sqrt{ \Big( 15\cos(100^\circ) - 7\cos(40^\circ)\Big)^2 + \Big( 15\sin(100^\circ) - 7\sin(40^\circ)\Big)^2 } \\ &=& \tiny{\sqrt{ 15^2\cos^2(100^\circ) -2*15*7\cos(100^\circ)\cos(40^\circ) + 7^2\cos^2(40^\circ) + 15^2\sin^2(100^\circ) -2*15*7\sin(100^\circ)\sin(40^\circ) + 7^2\sin^2(40^\circ) }} \\ &=& \tiny{\sqrt{ 7^2\Big(\underbrace{\sin^2(40^\circ)+\cos^2(40^\circ)}_{\small{=1}}\Big) + 15^2\Big(\underbrace{\sin^2(100^\circ)+\cos^2(100^\circ)}_{\small{=1}}\Big) -2*15*7\Big(\underbrace{\cos(100^\circ)\cos(40^\circ) + \sin(100^\circ)\sin(40^\circ)}_{\small{=\cos(100^\circ-40^\circ)}} \Big) }} \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(100^\circ-40^\circ) } \\ &=& \sqrt{ 7^2+ 15^2 -2*15*7\cos(60^\circ) } \quad | \quad \cos(60^\circ)= \dfrac{1}{2} \\ &=& \sqrt{ 7^2+ 15^2 - 15*7 } \\ &=& \sqrt{ 169 } \\ &=& \mathbf{13} \\ \hline \end{array}\)

The length of the line is 13

∑[(1E198 - (2n - 1)), n, 1, 199] =1E200 =100^100

2.

Find the remainder when r^14 is divided by r - 1.

The remainder is 1

1.
When \(x^4 + 6x^3 - ax^2 - 45x -15\) is divided by \(x^2 - x - 6\) the remainder is \(3\).
What is \(a\)?

\(\text{Let $x^2 - x - 6 = (x-3)(x+2)$}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{x^4 + 6x^3 - ax^2 - 45x -15}{(x-3)(x+2)}} &=& \mathbf{q(x) + \dfrac{3}{(x-3)(x+2)}} \quad | \quad \times (x-3)(x+2) \\\\ x=3: & x^4 + 6x^3 - ax^2 - 45x -15 &=& q(x)(x-3)(x+2) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& q(x)(0)(5) + 3 \\ & 3^4+6(3^3)-a(3^2)-45(3)-15 &=& 3 \\ & 81+162-9a-135-15 &=& 3 \\ & 93-9a &=& 3 \quad | \quad :3 \\ & 31-3a &=& 1 \\ & 3a &=& 31-1 \\ & 3a &=& 30 \quad | \quad :3 \\ & \mathbf{a} &=& \mathbf{10} \\ \hline \end{array}\)