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lakkis [162]
3 years ago
7

Kelly has a muffin recipe that calls for 1 1/2 cups of sugar. She wants to use 1/2 that amount of sugar. How much sugar will she

use
Mathematics
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

The correct answer would 3/4 cups of sugar

Step-by-step explanation:

divide 1 1/2 by 2 to get 3/4

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Simplify the expression <br>-6u - 9u + -9 + 5 + -9u​
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Answer:

-24u-4

Step-by-step explanation:

Combine like terms

-6u-9u-9u + -9+5

-24u-4

If you wanna factor it you can do that:

-4(6u+1)

5 0
3 years ago
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4p + 3= 22.6

There hope this helps you

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3 years ago
How do i graph y=1/2Sinø/2?
podryga [215]
\fb \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\&#10;% function transformations for trigonometric functions&#10;\begin{array}{rllll}&#10;% left side templates&#10;f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}&#10;\\\\&#10;f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\&#10;f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}&#10;\end{array}&#10;\\\\&#10;-------------------

\bf \bullet \textit{ stretches or shrinks}\\&#10;\left. \qquad   \right. \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\&#10;\left. \qquad   \right. \textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\&#10;\left. \qquad   \right. \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}

\bf \bullet \textit{function period or frequency}\\&#10;\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

now, with that template in mind,

\bf \stackrel{parent~function}{y=sin(\theta )}\qquad \qquad y=\stackrel{A}{\frac{1}{2}}sin\left(\stackrel{B}{\frac{1}{2}}\theta   \right)&#10;\\\\\\&#10;Amplitude\implies \frac{1}{2}&#10;\\\\\\&#10;Period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{1}{2}}\implies 4\pi

which is pretty much the same sin(θ) function, but squished by 1/2 and elongated up to 4π, check the picture below.


7 0
3 years ago
There's got to be someone who can answer this
Sphinxa [80]
Yes based on ja 82 and it can be added
4 0
3 years ago
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