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3 years ago

7

Relative to the circle with the equation x^2 + y^2 = 4 how has the circle with the equation (x+5)^2 + (y-6)^2 = 4 been shifted,

what is the radius of the circle

2 answers:

jolli1 [7]3 years ago

5 0

Math is so not my thing its so hard

mixas84 [53]3 years ago

3 0

Answer:

the circle is shifted -5 on the x and +6 on the y. Center is (-5,6). Radius = 2.

Step-by-step explanation:

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Naika took out a $29,300 5 year Federal Loan with an interest rate of 3.22%. She will

mina [271]

9514 1404 393

Answer:

$4717.30

Step-by-step explanation:

Assuming the interest is capitalized only at the start of repayment, the accrued interest is ...

I = Prt

I = ($29,300)(0.0322)(4 +1) = $4717.30

The accrued interest is $4717.30.

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3 years ago

Absolute value of −4.5

ddd [48]

Answer:

<h2>|-4.5| = 4.5</h2>

Step-by-step explanation:

|a| = a for a ≥ 0

|a| = -a for a < 0

therefore

|-4.5| = -(-4.5) = 4.5

4 0

2 years ago

Randomly distributing 45 pencils and 35 markers to his students. what is the probability that the first item he hands out will

Dominik [7]

If he hands out pencils will be 45/80 =
If he hands out markers will be 35/80

3 0

2 years ago

Find a difference 118.419 - 6.74

nevsk [136]

Step-by-step explanation:

The difference between 118.419 and 6.74 is

= 111.679

Hope it helps

7 0

3 years ago

Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g

SpyIntel [72]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

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

with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad B
\end{array}$

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.

8 0

2 years ago

Read 2 more answers