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3 years ago

Express 8.21 as a mixed number.

Mathematics

2 answers:

andrew-mc [135]3 years ago

6 0

$8 \frac{21}{100}$ will be the mixed number for the decimal value 8.21

geniusboy [140]3 years ago

3 0

I think that it is 8 21/100

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72 ounces in 6 steaks

svet-max [94.6K]

12 ounces for each steak ..............

6 0

3 years ago

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ASE

eimsori [14]

Answer:

128 deg

Step-by-step explanation:

The sum of the measures of the angles of a polygon of n sides is

(n - 2)180

A trapezium has 4 sides, so the sum of the measures of the angles is

(4 - 2)180 = 2(180) = 360

m<CBE + m<C + m<D + m<DEB = 360

Angles ABC and CBE are a linear pair so the sum of their measures is 180 deg.

m<ABC + m<CBE = 180

115 + m<CBE = 180

m<CBE = 65

m<C = x

m<D = 90

Angles DEB and DEF are a linear pair so the sum of their measures is 180 deg.

m<DEB + m<DEF = 180

m<DEB + m<103 = 180

m<DEB = 77

m<CBE + m<C + m<D + m<DEB = 360

65 + x + 90 + 77 = 360

x + 232 = 360

x = 128

Answer: 128 deg

3 0

3 years ago

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The mass of a basket containing 3 balls was 3.64 kilograms. The mass of the same basket and 5 balls is 5.26 kilograms. Find the

hram777 [196]

Answer:

1.21 kilograms

Step-by-step explanation:

Let y represent the mass of the empty basket, and x represent the mass of each ball

Create a system of equations:

3x + y = 3.64

5x + y = 5.26

Solve by elimination by multiplying the top equation by -1

-3x - y = -3.64

5x + y = 5.26

Add them together, then solve for x:

2x = 1.62

x = 0.81

Plug this back into one of the equations and solve for y:

3x + y = 3.64

3(0.81) + y = 3.64

2.43 + y = 3.64

y = 1.21

So, the mass of the empty basket is 1.21 kilograms

4 0

3 years ago

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Write an equation of cosine function with amplitude 2 and a period 4pi

Arisa [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}
\end{array}$

$\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{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that

$\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}$

8 0

4 years ago

License plate numbers in a certain state consists of seven characters. The first character is a digit (0 through 9). The next fo

Yuri [45]

Answer:

67,600

Step-by-step explanation:

For the first and last two characters, there are ten possible digits (0 through 9) for each character.

For the second through sixth character, there are 26 possible letters (A through Z) for each character.

The number of all possible license plate combinations "n" is given by the multiplication of each digits range of possible outcomes:

$n=10*26*26*26*26*10*10\\n=456,976,000$

Since there are no restrictions for repeated characters, there are 456,976,000 different license plate numbers are possible.

4 0

3 years ago