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3 years ago

1.

Mathematics

1 answer:

marishachu [46]3 years ago

6 0

Answer:

40%

Step-by-step explanation:

Well, 6/15 of the pieces of fruit are apples, so all you have to do is divide 6 by 15 to give you 0.4. This means you have 0.4 out of 1, which, if you multiply by 100, would give you 40 percent.

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9.21 <br><img src="https://tex.z-dn.net/?f=9.21%20%5Cdiv%200.4%20%3D%20" id="TexFormula1" title="9.21 \div 0.4 = " alt="9.21 \di

Whitepunk [10]

Let us do it by converting the 0.4 to a fraction, it has one decimal, so we'll use one zero at the denominator.

and 9.21 to a fraction as well, two decimals, thus two zeros at the denominator then,

$\bf 9.21\div 0.4\\\\
-------------------------------\\\\
0.\underline{4}\implies \cfrac{04}{1\underline{0}}\implies \cfrac{4}{10}\implies \cfrac{2}{5}\qquad \qquad \qquad \qquad 9.\underline{21}\implies \cfrac{921}{1\underline{00}}\\\\
-------------------------------\\\\
9.21\div 0.4\implies \cfrac{921}{100}\div \cfrac{2}{5}\implies \cfrac{921}{100}\cdot \cfrac{5}{2}\implies \cfrac{921}{20}\cdot \cfrac{1}{2}\implies \cfrac{921}{40}
\\\\\\
23\frac{1}{40}\implies 23.025$

3 0

3 years ago

Which is the solution set of the inequality 15 y minus 9 less-than 36 y greater-than nine-fifths y less-than nine-fifths y less-

Vaselesa [24]

Answer:

C. y<3

Step-by-step explanation:

We are required to determine the solution set of the inequality

$15y-9$

Step 1: Add 9 to both sides

$15y-9+9$

Step 2: Divide both sides by 15

$15y \div 15$

The solution set of the given inequality is: y<3.

The correct option is C.

4 0

3 years ago

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Molly is wrapping up part of sandwiches to sell at her sandwich cart. She cuts each sandwich in fourths and then wrap each fourt

Naily [24]

There fractions not holes

8 0

2 years ago

What's 285÷7=<br> What's the answer now

OverLord2011 [107]

40.71428571428571
If you want you can round it

5 0

3 years ago

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What are the period and phase shift for f(x) = −4 tan(x − π)?

nadya68 [22]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% function transformations for trigonometric functions
% templates
f(x)=Asin(Bx+C)+D
\\\\
f(x)=Acos(Bx+C)+D\\\\
f(x)=Atan(Bx+C)+D
\\\\
-------------------$

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

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{C}{B}\\
~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{C}{B}\textit{ is positive, to the left}\\\\
\bullet \textit{vertical shift by }D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}$

$\bf \bullet \textit{function period or frequency}\\
~~~~~~\frac{2\pi }{B}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
~~~~~~\frac{\pi }{B}\ for\ tan(\theta),\ cot(\theta)$

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

$\bf f(x)=\stackrel{A}{-4}tan(\stackrel{B}{1}x\stackrel{C}{-\pi })+\stackrel{D}{0}
\\\\\\
\stackrel{period}{\cfrac{2\pi }{B}\implies \cfrac{2\pi }{1}\implies 2\pi }
\\\\\\
\stackrel{phase/horizontal~shift}{\cfrac{C}{B}\implies \cfrac{-\pi }{1}}\implies -\pi \qquad \textit{shifted to the right by }\pi \textit{ units}$

3 0

3 years ago

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