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3 years ago

What are the steps used to transform the general form of the equation of a circle to standard form

Mathematics

1 answer:

Bezzdna [24]3 years ago

4 0

Answer:

Step-by-step explanation:

Convert the equation of a circle in general form shown below into standard form. Find the center and radius of the circle. Group the x 's and y 's together. Consider the x2 and x terms only. Complete the square on these terms. Replace the x2 and x terms with a squared bracket.

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Use the inverse sine ratio when you know the length of a leg opposite the required angle. True or false

brilliants [131]

Answer:

False

Step-by-step explanation:

The function you use to find the angle will depend on the other information given. Regardless, finding the angle may involve the inverse sine function, not the inverse sine ratio.

_____

If you know the leg adjacent as well as the leg opposite, the inverse tangent function will be an appropriate choice.

5 0

4 years ago

Find the value of x help me pls

goldenfox [79]

Answer:

x=12

Step-by-step explanation:

Since the angle needs to add up to 90, you will first need to subtract 30 from 90. This will leave you with 60. Next you will divide 60 by 5 which will give you 12 which is the answer.

6 0

3 years ago

Read 2 more answers

Please help. linear equations.

postnew [5]

Answer:

wow

Step-by-step explanation:

wkanda might never be forever which means that still could be quite ttue?

8 0

3 years ago

You need half a liter of milk how many millimeters do you need

never [62]

Proably about 500 milleleters

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3 years ago

Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

$\bf \begin{array}{llll}
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

8 0

3 years ago