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densk [106]
2 years ago
14

When calculating the circumference of a circle, sometimes the radius is given instead of the diameter. What is the relationship

between radius and diameter? a. radius is twice the diameter c. diameter is twice the radius b. diameter is half the radius d. no relationship, can't find circumference with radius Please select the best answer from the choices provided A B C D
Mathematics
2 answers:
marin [14]2 years ago
8 0

Answer:

Option A ( radius is twice the daimeter )

Step-by-step explanation:

The relationship between radius and daimeter is given as

       Radius² = Daimeter    ⇒    r² =  d

And cricumference can be calculated with radius

 As circumference is    2πr

UkoKoshka [18]2 years ago
8 0

Answer:  c. diameter is twice the radius

For example, if the radius is r = 10 feet then the diameter is twice that and it's d = 2*r = 2*10 = 20 feet.

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Given the frequency table, what percentage of the students that like rock are also in grades 11–12? Round to the nearest whole p
Ostrovityanka [42]

Answer:

d. 45%

Step-by-step explanation:

In total there are 30 + 25 = 55 students in total that like rock. From these there are 25 who are in grades 11-12. This makes that (25/55) * 100% = 45.45% of the students that like rock are in grades 11-12.

7 0
3 years ago
Read 2 more answers
What kind of shift does the graph have? y = -sin (x-1) from y = sin(x)
Mademuasel [1]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\end{array}
\\\\
-------------------\\\\

\bf \bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\
\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}\\\\

\bf \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{function period or frequency}\\
\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

now, with that template in mind, let's see

\bf \begin{array}{lllll}
y=&-1sin(&1x&-1)&+0\\
&A&B&C&D
\end{array}
\\\\\\
\textit{horizontal shift of }\cfrac{C}{B}\implies \cfrac{-1}{1}\implies -1
\\\\\\
\textit{A is negative, so is flipped upside-down}
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3 years ago
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Answer:

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Step-by-step explanation:

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vfiekz [6]
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