All Answers 356458 Answers

We  want to  find   where  the y  value  of the  graph =   -1

Notice  that  the  graph intersects  the  line  y =  -1    at  4 places

So.....4  values of c   where this occurs

Hope  that makes sense  

I think you would do 5!(4) or something to that effect.

Such a triangle is not possible

By the triangle inequality :

XY  +  XZ  must  be greater  than   YZ

But

25  +  25    =   50   =    YZ

(2a + b) - (a + b) = 1 - 2 (mod 10)

a = 9 (mod 10)

a + b = 2 (mod 10)

9 + b = 2 (mod 10)

b = 3

a = 9, b = 3, a - b = 6

=^._.^=

She will go up if she gets more heads than tails, and she will go down if she gets more tails than heads. 

If when stops at the 7th floor or higher, that means she went up.

Since there's an equal chance of getting more heads than tails and getting more tails than heads, the answer is 1/2.

=^._.^=

-6t^2  + 51t =  118

-6t^2  +  51t   - 118 =   0                     multiply through by   -1

6t^2   - 51t  +  118  =   0

The discriminant  =      51^2  - 4*6 * 118    =   -231

So......this never occurs....see  the  graph here  :  https://www.desmos.com/calculator/nbovxklqyd

The max height reached is ≈  108 ft

3+7 = 10 and that's wrong

1+1=2

Think of it this way if you have one apple and you add another apple you know  have 2 apples.(:

If we write \(\sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) }\)
in the form \(\dfrac{(a*\sqrt{2} + b*\sqrt{3})}{c}\) such that a, b, and c

are positive integers and c is as small as possible,
then what is a + b + c?

\(\begin{array}{|rcll|} \hline && \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } * \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} + 3*\sqrt{3})(2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (4*2-9*3) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ -19 } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{-(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*(\sqrt{2} + \sqrt{3}) -(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*\sqrt{2} + 19*\sqrt{3} -2*\sqrt{2} + 3*\sqrt{3}}{ 19 } \\\\ &=& \dfrac{{\color{red}17}*\sqrt{2} + {\color{red}22}*\sqrt{3}}{ {\color{red}19} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+b+c &=& 17+22+19 \\ \mathbf{a+b+c} &=& \mathbf{58} \\ \hline \end{array}\)

l^2  + (l-9)^2  = (l+9)^2

l^2 + l^2 -18l +81 = l^2 + 18l + 81

l^2 - 36l = 0

l ( l-36) = 0

l = 0   throw out       l = 36      other leg is  27      hyp = 45

The minimum length of CQ = 3(5 - √17)

It seems you have copied a test question .....

You just keep substituting

f(-1)=1+2=3

f(f(-1))=f(3)=9-6=3

since f(3)=3 the answer is not going to change from here so the answer is 3

Like this

\(\lfloor 2x^2-x \rfloor=0\\ \text{It will be true when }\qquad 0\le2x^2-x <1\\~\\ Let\;\;y=2x^2-x\\ \text{It will be true when }\qquad 0\le y<1\\~\\ \)

y=2x^2-x \\

y=x(2x-1)

this is a concave up parabola with roots of x=0 and x=0.5

So between x=0 and x=0.5 the y value will be negtive so those points are no good.

Where does this parabola intersect with y=1?

\(y=1\\ 1=2x^2-x\\ 2x^2-x-1=0\\ \)

I solved using the quadratic equation
\(x=1 \;\;or\;\;-0.5\\~\\ \text{So the statement will be true for }\quad \\~\\ -0.5

If

\(d-6c=4\) and \(2d-9c=20\), find the value of \(\dfrac{d}{c}\).

\(\begin{array}{|rcll|} \hline 2d-9c &=& 20 \quad | \quad :c \\\\ \mathbf{\dfrac{2d}{c}-9} &=& \mathbf{\dfrac{20}{c}} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline d-6c &=& 4 \quad | \quad :c \\\\ \dfrac{d}{c}-6 &=& \dfrac{4}{c} \quad | \quad *5 \\\\ \mathbf{\dfrac{5d}{c}-30} &=& \mathbf{\dfrac{20}{c}} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (2)=(1): & \dfrac{5d}{c}-30 &=& \dfrac{2d}{c}-9 \\\\ & \dfrac{3d}{c} &=& 30 -9 \\\\ & \dfrac{3d}{c} &=& 21 \quad | \quad :3 \\\\ & \mathbf{\dfrac{d}{c}} &=& \mathbf{7} \\ \hline \end{array}\)

...In this case, it’s just a brain-dead head.

Dead Head ... I was wondering if he is Grateful, but ...

Given that x satisfies \(x^2 - 5x + 1 = 0\),
find the value of \(x^4 +\dfrac{1}{ x^4 }\).

\(\begin{array}{|rcll|} \hline \mathbf{x^2 - 5x + 1} &=& \mathbf{0} \\\\ x_{1,2} &=& \dfrac{5\pm\sqrt{25-4}}{2} \\ x_{1,2} &=& \dfrac{5\pm\sqrt{21}}{2} \\ \hline x_{1,2}^2 &=& \left( \dfrac{5\pm\sqrt{21}}{2} \right)^2 \\ x_{1,2}^2 &=& \dfrac{123\pm 5\sqrt{21}}{2}\\ \hline x_{1,2}^4 &=& \left( \dfrac{123\pm 5\sqrt{21}}{2} \right)^2 \\ x_{1,2}^4 &=& \dfrac{527\pm 115\sqrt{21}}{2} \\ \hline \end{array}\)

\(x_1:\)

\(\begin{array}{|rcll|} \hline x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{527+115\sqrt{21}} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{(527+115\sqrt{21})}*\dfrac{(527-115\sqrt{21})}{(527-115\sqrt{21})} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2(527-115\sqrt{21})}{4} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{(527-115\sqrt{21})}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}+527-115\sqrt{21}}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{2*527}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& 527 \\ \hline \end{array}\)

\(x_2:\)

\(\begin{array}{|rcll|} \hline x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{527-115\sqrt{21}} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{(527-115\sqrt{21})}*\dfrac{(527+115\sqrt{21})}{(527+115\sqrt{21})} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2(527+115\sqrt{21})}{4} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{(527+115\sqrt{21})}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}+527+115\sqrt{21}}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{2*527}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& 527 \\ \hline \end{array}\)

\(x^4 +\dfrac{1}{ x^4 } =527\)

3T  + 2C  =   21     ⇒    6T  + 4C  =      42

2T  + 3C  =  19     ⇒    -6T  -  9C  =   - 57

Add the last  two  equations

-5C  =  - 15

C  =   -15  / -5     =    3

So

3C  =  3 (3) =   9

Just add them together , Dead-Head.....

it's -2