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larisa [96]
4 years ago
7

Need help all mult choice show some type of work/proof

Mathematics
1 answer:
Brrunno [24]4 years ago
6 0

AD is the diagonal of a 6 by 15 rectangle,

31.

AD^2 = 6^2+15^2=36+225=261 = 9(29)

AD=3 \sqrt{29}

Choice A

32.

d^2 = 2^2+3^2=4+9=13

d=\sqrt{13}

Choice A

33.

7 - 4x = 7y

4x + 7y = -7

The parallel through (2,0) keeps the coefficients on x and y and just changes the constant:

4x + 7y = 4(2)+7(0)= 8

7y = -4x + 8

y = (-4/7) x + 8/7

Choice B

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Kay [80]

Answer:

-1.2 that is the point hope this helps

Step-by-step explanation:


5 0
3 years ago
This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

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

  • Multiply row 1 by 1/3:

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

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

6 0
4 years ago
The value of a 2 in the tens place is what fraction of the value of a 2 in the hundreds place?
lapo4ka [179]

Answer:

2/10

Step-by-step explanation:

it identifed it for you in the 10's place

3 0
3 years ago
I need help with number 5
Sergio [31]

Answer:

B

Step-by-step explanation:

This balances out the equation so both sides are equal

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3 years ago
Solve for x. Assume that lines which appear to be tangent are tangent.
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