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Speed = distance/time

so

Speed * time = distance

\(\frac{d_1}{10}+\frac{d_2}{4}+\frac{d_3}{20}=2.5\\~\\ \frac{d_1}{7.5}+\frac{d_2}{6}+\frac{d_3}{15}=3\\~\\ \frac{d_1}{15}+\frac{d_2}{3}+\frac{d_3}{40}=1.75\\~\\\)

\(\begin{bmatrix} \frac{1}{10} && \frac{1}{4} && \frac{1}{20} \\ \frac{1}{7.5} && \frac{1}{6} && \frac{1}{15} \\ \frac{1}{15} && \frac{1}{3} && \frac{1}{40} \\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2\\ d_3 \end{bmatrix}= \begin{bmatrix} 2.5 \\ 3\\ 1.75 \end{bmatrix} \)

\(\begin{bmatrix} 2 && 5 && 1 \\ 4 && 5 &&2\\ 8 && 40 && 3 \\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2\\ d_3 \end{bmatrix}= \begin{bmatrix} 50 \\ 90\\ 210 \end{bmatrix}\)

\(\begin{bmatrix} 8 && 20 && 4 \\ 8 &&10 &&4\\ 8 && 40 && 3 \\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2\\ d_3 \end{bmatrix}= \begin{bmatrix} 200 \\ 180\\ 210 \end{bmatrix}\)

\(\begin{bmatrix} 8 && 20 && 4 \\ 0 &&-20 &&0\\ 0 && 20 && -1 \\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2\\ d_3 \end{bmatrix}= \begin{bmatrix} 200 \\ -20\\ 10 \end{bmatrix}\)

\(..............\\~\\ \begin{bmatrix} 2 && 5 && 1 \\ 0 &&1 &&0\\ 0 && 0 && 1\\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2\\ d_3 \end{bmatrix}= \begin{bmatrix} 50 \\ 2\\ 30 \end{bmatrix}\)

\(d_3=30\qquad \\ d_2=2\\ d_1=5\)

Run 5miles,       swim 2 miles      and        cycle  30miles

Coding:

\frac{d_1}{10}+\frac{d_2}{4}+\frac{d_3}{20}=2.5\\~\\

\frac{d_1}{7.5}+\frac{d_2}{6}+\frac{d_3}{15}=3\\~\\

\frac{d_1}{15}+\frac{d_2}{3}+\frac{d_3}{40}=1.75\\~\\

begin{bmatrix}
\frac{1}{10} && \frac{1}{4}  && \frac{1}{20} \\
\frac{1}{7.5} && \frac{1}{6}  && \frac{1}{15} \\
\frac{1}{15} && \frac{1}{3}  && \frac{1}{40} \\
\end{bmatrix}

begin{bmatrix}
d_1  \\
d_2\\
d_3
\end{bmatrix}=
begin{bmatrix}
2.5  \\
3\\
1.75
\end{bmatrix}

begin{bmatrix}
2 && 5  && 1 \\
4 && 5  &&2\\
8 && 40  && 3 \\
\end{bmatrix}
begin{bmatrix}
d_1  \\
d_2\\
d_3
\end{bmatrix}=
begin{bmatrix}
50  \\
90\\
210
\end{bmatrix}

begin{bmatrix}
8 && 20  && 4 \\
8 &&10  &&4\\
8 && 40  && 3 \\
\end{bmatrix}
begin{bmatrix}

begin{bmatrix}
8 && 20  && 4 \\
0 &&-20  &&0\\
0 && 20  && -1 \\
\end{bmatrix}

begin{bmatrix}
d_1  \\
d_2\\
d_3
\end{bmatrix}=
begin{bmatrix}
200  \\
-20\\
10
\end{bmatrix}
d_1  \\
d_2\\
d_3
\end{bmatrix}=
begin{bmatrix}
200  \\
180\\
210
\end{bmatrix}

..............\\~\\
begin{bmatrix}
2 && 5  && 1 \\
0 &&1  &&0\\
0 && 0  && 1\\
\end{bmatrix}
begin{bmatrix}
d_1  \\
d_2\\
d_3
\end{bmatrix}=
begin{bmatrix}
50  \\
2\\
30
\end{bmatrix}

[DEF] / [ABC] = 37 / 80

Solve (2x + 4)/(x^2 + 4x - 5) = (2 - x)/(x + 5).

Hello Guest!

\(x\in \{\frac{1}{2}(1-i\sqrt{23}),\frac{1}{2}(1+i\sqrt{23})\}\)  [calculated with WolframAlpha]

  !

XY = 1/2(AB + CD) - AB 

https://web2.0calc.com/questions/trapezoid_25

1/n^2 isn't correct 

p/q = 18.

The least amount of paint is 25.6 gallons.

By Cauchy-Schwarz, the maximum value is 1/n^2.

The red line below is   x = 2       f(2)  is the 'y' value of the point  where the red line crosses the graph.....

Matt, Justin, Tony, and Gabriel shared $168. Gabriel received 1/7 of the amount of money received by Matt, Tony, and Justin. Matt received 3/4 of the amount of money received by Tony and Justin. Justin received 2/5 as much money as Tony. How much did Justin receive?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

M + J + T + G = 168

M+J+T => X

X + X/7 = 168             X = 147

Gabriel received =>    168 - 147 = 21

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

M + J + T = 147

J + T => Y

Y + 3Y/4 = 147           Y = 84

Matt received =>     147 - 84 = 63

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

J + T = 84

T + 2T/5 = 84            T = 60

Justin received =>     84 - 60 = 24   

When the rock hits the lunar surface   h = height = 0

0  = −t^2+ 18t + 10     Use Quadratic Formula to solve for t =      throw out negative value      a = -1    b = 18      c = 10

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Combine lie terms to get to ax^2 + bx + c form      min then occurs at   x =  - b/2a

                      use this value of x in the equation to calculate the function's minmum value

i got the answers! its 28

Volume =  L W H = 96  in³

             (8+h)(h-2)(h) = 96

              h³-6h² -16h-96=0

        h = 4    via   online algebra calculator            h = 4    l = 12    w = 2 inches

If the first term of an arithmetic sequence equals the third minus the second, and the fourth term is 18, what is the first term?

Hello Guest!

  a = c  - b

  d = 18

a - a - b - a - c - a - 18

 18 - 3a = a

  18 = 4a

   a = 4.5

 a,    b,    c,     d

4.5,  9,  13.5, 18

  !

well

x^2-16 must be greater or equal to 0

and

sqrt{x^2-16) must be greater or equal to 3

So solve both of those and you will have the domain.

       N D Q

1  = 5 0 1
2  = 3 1 1
3  = 1 2 1
4  = 0 0 2

You can make change for 50 cents in 4 different ways.

If c = -1/2       the    10x^4   in g(x) will become -5 x^4    and will cancel the   5x^4 in f(x)   leaving a polynomial of degree 3 when the f(x) and c (g(x) are added.

After Marcus gave away 1/4 of his stamps, he still has (3/4)x.

After Kevin gave away 1/3 of his stamps, he still has (2/3)x.

(3/4)x - (2/3)x = 15               Multiply by 12

9x - 8x = 180

so x = 180. 

The problem wants us to find (1/4)x. (1/4)x = (1/4)*180 = 45.

Easy

let d be ducks and c be cows

we can form 2 equations: d + c = 24 and 2d + 4c = 66. using the first equation, we can get 2d + 2c = 48. subtracting the 2 equations gives 2c = 18. so c = 9,  and d = 15. 

The domain is (-inf,-6] U [6,inf).