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qwelly [4]
2 years ago
11

Suppose sin a = 12/13 and sin b= 5/11. FInd cos(a+b)

Mathematics
1 answer:
m_a_m_a [10]2 years ago
6 0

Answer:

1.28 or 912/715

Step-by-step explanation:

sine= opposite side/hypotenuse

cos=adjacent/ hypotenuse

sine a= 12/13 using the Pythagorean Theorem and solve for the missing side. (which is the adjacent side)

cos a = 5/13

Do the same for sine b

cos b= 9.8/11

add both the cosine value

(5/13)+(9.8/11)=1.28 or 912/715 in fraction form

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(3.14)(21.98)(21.98)

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3 years ago
Total blood cholesterol level was measured for each of
laiz [17]
<h3>Answer: Mean = 218.9.</h3><h3>Median = 229</h3><h3>Mode = Zero mode.</h3>

Step-by-step explanation:

Given blood cholesterol level was measured for each of 8 adults (in mg/dL) are:

264, 191, 160, 148, 262, 212, 268, 246

In order to find the mean, we need to add all those 8 numbers and divide by 8.

Therefore, mean = \frac{264+191+160+148+262+212+268+246}{8} =\frac{1751}{8} =218.9.

<h3>Mean = 218.9.</h3>

In order to find the median, we need to arrange them in ascending order:

148, 160, 191, 212, 246, 262, 264, 268.

The middle most two values are 212 and  246.

Therefore, median = \frac{212+246}{2} =\frac{458}{2} = 229.

<h3>Median = 229.</h3>

\mathrm{The\:mode\:is\:the\:term\:in\:the\:data\:set\:that\:appears\:the\:most.}

\mathrm{If\:there\:is\:more\:than\:one\:term\:that\:appears\:the\:most,\:then\:there\:is\:no\:mode.}

\mathrm{Count\:the\:number\:of\:times\:each\:element\:appears\:in\:the\:list}

\begin{pmatrix}148&160&191&212&246&262&264&268\\ 1&1&1&1&1&1&1&1\end{pmatrix}

\mathrm{The\:most\:common\:element\:is\:not\:unique,\:so\:there\:is\:no\:mode}

<h3>Therefore, Mode = Zero mode.</h3>
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3 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

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which is a very small number close to zero.</span>
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jolli1 [7]

Answer:

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_____

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expeople1 [14]

Answer:

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3 years ago
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