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melomori [17]
3 years ago
11

X^2 = -8x - 7 Rewrite the equation by completing the square. What are the solutions?

Mathematics
1 answer:
gladu [14]3 years ago
6 0

1) rewrite the equation by completing the square

(X+4)^2=9

2) what are the solution to the equation

Answer A= X=-4 plus and minus 3

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Write down 7 numbers that have a range of 10 and a mean of 12.<br> Please show the working!!
posledela
7, 9, 10, 12, 14, 16, 17

You can find that the range is ten by subtracting the largest number (17) by the smallest number (7) which leaves you with 10. You can find the mean by crossing off a number from each side, one at a time, until you land on the middle which is 12. I hope this helps, let me know if you have any questions!
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NEED HELP ASAP! TYSM!
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Which graph best represents -5y = -6x + 15
frutty [35]

Answer: Option c.

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y=mx+b

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y = \frac{-6}{-5}x + \frac{15}{-5}\\\\y = \frac{6}{5}x -3

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m=\frac{6}{5}\\\\b=-3

A posive slope means that the the line moves upward from left to right.

With this information, you can conclude that the the graph that  best represents the given equation is the graph shown in the Option c.

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4 years ago
If AABC is reflected across the yaxis, what are the coordinates of A?
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(-1,3)

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3 years ago
Disregard my work, but how do you get the answer? There is also a graph at the bottom which is optional to use.
kakasveta [241]
\bf ~~~~~~~~~~~~\textit{function transformations}&#10;\\\\\\&#10;% templates&#10;f(x)=  A(  Bx+  C)+  D&#10;\\\\&#10;~~~~y=  A(  Bx+  C)+  D&#10;\\\\&#10;f(x)=  A\sqrt{  Bx+  C}+  D&#10;\\\\&#10;f(x)=  A(\mathbb{R})^{  Bx+  C}+  D&#10;\\\\&#10;f(x)=  A sin\left( B x+  C  \right)+  D&#10;\\\\&#10;--------------------

\bf \bullet \textit{ stretches or shrinks horizontally by  }   A\cdot   B\\\\&#10;\bullet \textit{ flips it upside-down if }  A\textit{ is negative}\\&#10;~~~~~~\textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }  B\textit{ is negative}\\&#10;~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{  C}{  B}\\&#10;~~~~~~if\ \frac{  C}{  B}\textit{ is negative, to the right}\\\\&#10;~~~~~~if\ \frac{  C}{  B}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{ vertical shift by }  D\\&#10;~~~~~~if\   D\textit{ is negative, downwards}\\\\&#10;~~~~~~if\   D\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{  B}

with that template in mind.

since the original or "parent" function is y = x²−4x+3, if we change the "x" argument to "x-2", we end up with a horizontal shift.

the x-2 part would be in the template the Bx+C part, with B = 1 and C = -2, or a horizontal shift to the right of 2/1 or 2 units.

since the parent function has a point at (2, -1), if we move that horizontally only to the right, we'd end up at (4, -1).
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