All Answers 358149 Answers

"Is" means " =  "

"Of " means  multiply

"What number"  ....let this be N

So we have

32 = 40% x N     convert 40% to a decimal

32 = .40 x N       divide both sides by .40

32 / .40 = N = 80

x+y=10 x-y=2 solve using elemination(allgerbra 2) ?

$$\\ \frac{(x+y)\textcolor[rgb]{1,0,0}{+}(x-y)}{2} = x = \frac{10+2}{2} = \frac{12}{2} = 6 \\
\frac{(x+y)\textcolor[rgb]{1,0,0}{-}(x-y)}{2} = y = \frac{10-2}{2} = \frac{8}{2} = 4$$

How did you get from

cos(15) - sin(15)    to

√2cos(15 + 45)   and then to

-√2sin(15 - 45)  ????

NOOOOOOOOOO!!!!!!!!!!

.......؟

sin 105 - sin 15

$$\sin{( 105\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\
=\sin{( 90\ensurement{^{\circ}} + 15\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\
= \underbrace{\sin{( 90\ensurement{^{\circ}} ) }}_{=1}\cos{( 15\ensurement{^{\circ}} ) }+\underbrace{\cos{( 90\ensurement{^{\circ}} ) }}_{=0} \sin{( 15 \ensurement{^{\circ}}) }- \sin{( 15\ensurement{^{\circ}} ) }\\
=\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} )} - \sin{( \textcolor[rgb]{1,0,0}{15 \ensurement{^{\circ}}}) } \\ \\
\boxed{
=\sqrt{2}\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} +45\ensurement{^{\circ}} )}\\
=-\sqrt{2}\sin{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} -45 \ensurement{^{\circ}})} \\
=\frac{\sqrt{2}}{2} = 0.70710678119
}$$

Ah...I see....my guess was wrong....thanks for that link....I'm going to look at that very closely...!!!

 Hi Chris,

The sum of the roots of a polynomial of degree n is   

$$\mbox{sum of the real roots }= -\frac{\mbox{coefficient of }x^{n-1}}{\mbox{coefficient of }x^n}$$

but there is a lot more to this Chris.

Have a look  at this

Melody...just playing around with this in WolframAlpha....it appears that in a polynomial of degree n, that the sum of the roots will always be the coefficient on the x^n-1 term, unless the sum of the roots = 0. In that case, that power (naturally) isn't represented in the polynomial. The ending constant will (again, naturally) be the product of the roots !!!

Ah.....that makes it much clearer.....!!!

Think of it like this:

(x - x₁)(x - x₂) = 0

x² - (x₁ + x₂)x +x₁x₂ = 0

.

.

Seconds in an hour = 3600

3600sec x 5ft/sec  = 18000 ft

Add 2 0s when you *100. I also want to ask you how many zeroes there are in that number you put in the title. Here is a link you can click on to see names of big numbers.

|-8x+5|≥ 10

We can solve two equations here, the first  is

-8x + 5 ≥ 10     subtract 5 from both sides

-8x ≥ 5             divide by -8 and reverse the inequality sign

x ≤ -5/8

The second equation is

- (-8x + 5) ≥ 10 

8x - 5 ≥ 10      add 5 to both sides

8x ≥ 15           divide both sides by 8

x ≥ 15/8

So the solutions are (-∞, -5/8] U [15/8, ∞ )

Thanks for the answer.

That would be a good answer. Besides, 1!=1.

Find the value of  x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and ( x, 11) is 15 square units.

$$area = \pm15 =
\dfrac{
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
}
{2}
\\\\
\pm30 =
\left|
\left[
\left(\begin{array}{c}
x\\
11
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\times\left[
\left(\begin{array}{c}
8\\
2
\end{array} \right)
-
\left(\begin{array}{c}
6\\
5
\end{array} \right)
\right]
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
11-5
\end{array} \right)
\times
\left(\begin{array}{c}
8-6\\
2-5
\end{array} \right)
\right|
} \\
\pm30 =
\left|
\left(\begin{array}{c}
x-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\\\
x_1 =?\\
30 =
\left|
\left(\begin{array}{c}
x_1-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_1-6)(-3)-6*2=30\\
(x_1-6)(-3)=42\\
(x_1-6)=-14\\
\textcolor[rgb]{1,0,0}{x_1 =-8}\\\\$$

$$x_2 =?\\
-30 =
\left|
\left(\begin{array}{c}
x_2-6\\
6\end{array} \right)
\times
\left(\begin{array}{c}
2\\
-3\end{array} \right)
\right|
} \\
(x_2-6)(-3)-6*2=-30\\
(x_2-6)*3+12=30\\
(x_2-6)*3=18\\
x_2-6=6\\
\textcolor[rgb]{1,0,0}{ x_2=12 }$$

a. is okay

Since I do not have time for a new puzzle, I'll post something which people can comment on. 

Steve flips a coin, 
If tail appears in the first flip, you receive $2, if tail appears in the second flip, you receive $4, if tail appears in the third flip, you receive $8, if tail appears in the fourth flip, you receive $16, etc. The money will keep on doubling until you hit a tail. 

How much would you be willing to pay to play this game?

I am not sure what is less and what is more. I would had paid less than normal because there is a tiny bit of chance on landing on its side.

A guy wire is attached to the top of a 75 m tower and meets the ground at a 65 degree angle. How long is the wire?

We can use the sine function to solve this

sin(65) = 75 / h      where h is the wire length ...  rearrange ......

h = 75 / sin (65) = about 82.75 ft

$$\small\text{
$
\sin{ ( 12^{12^{14}} \ensurement{^{\circ}} ) } \\
=\sin{ ( 12^{ 168 } \ensurement{^{\circ}} ) } \\
= \sin{ ( 216 \ensurement{^{\circ}} ) } \\
= -0.587785252292
$
}}$$

Hi Chris,

What Heureka has done has many implications.  it can be used on polynomials of degree higher than 2.

although I'd need to 'think' with it if I was asked a question with a higher degree.

I think it can be used in probablility too although again I am very rusty