Math Help plases I need an answer

Mathematics

1 answer:

Makovka662 [10]3 years ago

5 0

The answer would be A, since the median is 14 and that is where the middle line is at

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You make both rocking chairs and bar stools. You charge $100 for every rocking chair and $40 for every bar stool. Which expressi

Gelneren [198K]

100r+40b=income
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4 years ago

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statuscvo [17]

We do
how much added-howmuch subtracted

$\int\limits^5_0 {F(t)} \, dt$ - $\int\limits^5_0 {E(t)} \, dt$
or
$\int\limits^5_0 {F(t)-E(t)} \, dt$
so
first simlify F(t)-E(t) (if you want, not necicary if use calculator)
(t+7-ln(t+4))/(t+2)=F(t)-E(t)

evaluate
$\int\limits^5_0 \frac{t+7-ln(t+4)}{t+2} } \, dt$ ≈9.05584
about 9 pints

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3 years ago

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sasho [114]

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3 years ago

lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.

zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$

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3 years ago

What is the exact value of cos 105°?

sergij07 [2.7K]

Answer:

$\frac{-\sqrt{6} + \sqrt{2} }{4}$

Step-by-step explanation:

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3 years ago