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Trava [24]
3 years ago
12

Solve s/10=7 The solution is s=

Mathematics
1 answer:
Alexeev081 [22]3 years ago
3 0

Answer:

s=70

t=30

r=32

:D!!

10×7=70

6×5=30

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Pls help<br><br> Given: △ABC, CM⊥ AB, BC = 5, AB = 7<br> CA = 4 sqrt(2)<br> Find: CM
babymother [125]

Answer:

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2  = 32

BM^2 + CM^2         = 25

(7 - BM)^2 - BM^2 = 7               (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7    

Step 4

Simplify And Solve

49 - 14BM = 7              Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42              Divide by - 14

BM = -42 / - 14

BM = 3

Step  Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2        Subtract 3^2 from both sides.

CM^2 = 25 - 9            

CM^2 = 16                     Take the square root of both sides.        

sqrt(CM^2) = sqrt(16)

CM = 4    < Answer

Step-by-step explanation:

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Use matrices and elementary row to solve the following system:
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I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

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Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

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3 years ago
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