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3 years ago

Find the area of the shaded region. Use 3.14 for π.

Mathematics

1 answer:

Lina20 [59]3 years ago

6 0

I hope this helps you

small circle area=3,14.19^2=1133,5

large circle area =3,14.24^2=1808,6

shaded area=1808,6-1133,5

shaded area=675,1

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For a function p(x), as |x|—> infinity then p(x)—> 3. Which of the following graphs could be p(x)?

VladimirAG [237]

Answer:

Step-by-step explanation:

A pe x the answer is d

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3 years ago

In a group of 60 students, 12 students take algebra one, 18 students take algebra 2 and 8 students take both subjects how many s

hoa [83]

60-12-18-2-8=20
=48-18-2-8
=30-2-8
=28-8
=20

So, 20 students don't take either of these subjects.

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3 years ago

A tetrahedron is a triangular pyramid with 4 equilateral faces. Han has a box of tetrahedrons with each of the 4 vertices marked

Daniel [21]

Theoretically, the number of the vertex comes at the top must be equal for infinitely many numbers of trials, but it's a tedious job to roll the dice for a large number of times. Rolling the tetrahedron only 10 times will not predict the correct result.

Han suspects that this tetrahedron is weighted in some way to make 1 appear more often than the other numbers.

So, check his suspect experimentally with the help of the concept of center of mass. A fair tetrahedron must have the center of mass at the center of the body.

The given tetrahedron has 4 equilateral triangular faces, so at first, measure all the sides of triangular faces and make sure that all are having the same length.

Then, conduct an experiment to check the location of the center of mass.

Hang the tetrahedron by attaching a thin string at the vertex 1 as shown in the figure and observe the base surface ( triangle 234) whether it is parallel to the ground or not.

Now, repeat the same by hanging it with vertices 2, 3, and 4 and observe the base surface.

For the center of mass to be located at the center of the tetrahedron, the base must be parallel to the ground (flat ground).i.e the hanging string must be perpendicular to the base for all the four cases.

If this is so, then the tetrahedron is a fair tetrahedron otherwise not.

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3 years ago

How do you find the scale factor?

Anika [276]

Answer:

The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure.

Step-by-step explanation:

6 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\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}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\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, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

3 years ago