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DaniilM [7]
3 years ago
12

Carl is boarding a plane. He has 2 checked bags of equal weight and a backpack that weighs 4 kg. The total weight of Carl's bagg

age is 35kg
Write an equation to determine the weight, w, of each of Carl's checked bags.
And the weight of each of Carl's check bags
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

(35 - 4)/2 = w and 15.5kg

Step-by-step explanation:

To get the weight of both bags you would have to first subtract the backpack's weight from the total weight. 35 - 4 = 31. Now divide by 2 to get the weight of one bag. 31/2 = 15.5. Each bag weighs 15.5 kg

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Read 2 more answers
What is the period and midline?
OverLord2011 [107]
\bf \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}}cos({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;&#10;\end{array}\qquad

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\&#10;&#10;&#10;\end{array}

\bf \begin{array}{llll}&#10;\bullet \textit{function period}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}&#10;

so if you notice yours \bf \begin{array}{llll}&#10;3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\&#10;&\ \uparrow&\uparrow \\&#10;&B&D &#10;\end{array}

now.. normally the function \bf 3.2cos&\left( \frac{5}{3}\theta \right)
 has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline  up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is 2\pi
so, the new period will be \bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0
3 years ago
Does anyone know how to do this
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How to do what?????????
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3 years ago
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