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Anni [7]
3 years ago
15

A circle is centered at the point (-3, 2) and passes through the point (1, 5). The radius of the circle is ? units. The point (-

7, ?) lies on this circle.
Mathematics
2 answers:
Llana [10]3 years ago
8 0

Answer:

radius 5. (-7,-1) is on the circle

%100 correct on PLATO


Step-by-step explanation:


olganol [36]3 years ago
3 0
Hello,

The circle has like equation: (x+3)²+(y-2)²=r²

If x=1 then y=5 ==>(1+3)²+(5-2)²=r²==>4²+3²=r²==>r=5 (the radius)

if x=-7 then (-7+3)²+(y-2)²=25 ==>  (y-2)²=25-16==>y=2+3 or y=2-3
The points (-7,5) and (-7,-1) are on the circle.
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Answer:

Step-by-step explanation:

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3 0
3 years ago
I really really really really really really really really need help
Effectus [21]

The answer would be 3.8.

So, to get 3 to 4, you would have to multiply 3 by something to get to 4. So you would divide 4 by 3 and get 1.3. So then you would divide 5 by 1.3 to solve for k, since k times 1.3 equals 5.

5 ÷ 1.3 = 3.8.

4 0
3 years ago
Find an equation in slope-intercept form of the line that has slope –9 and passes through point A(-9, - 1).
mojhsa [17]

Answer:

Step-by-step explanation:

y + 1 = -9(x + 9)

y + 1 = -9x - 81

y = -9x - 82

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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
3 years ago
17/18 - 8/9 equal out to
Nutka1998 [239]
\frac{17}{18}-\frac{8}{9}=x
You must find a common denominator(bottom)
\frac{8}{9}×\frac{2}{2}=\frac{16}{18}
\frac{17}{18}-\frac{16}{18}=\frac{1}{18}
Your answer is x=\frac{1}{18}
8 0
3 years ago
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