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3 years ago

15

Gabe filled 5 glasses of the same size with ice tea from a full 64-ounce pitcher. There were 3 ounces left in the pitcher. Which

equation and solution give the number of ounces, s, in each glass?

2 answers:

Helga [31]3 years ago

6 0

Answer:

The number of ounces in each glass is $12.2\ ounces$

Step-by-step explanation:

Let

s-------> the number of ounces in each glass

we know that

To get the number of ounces in each glass, subtract the amount of tea left in the pitcher from the total amount of tea and divide the result by the number of glasses.

so

The equation that represent the situation is

$s=(64-3)/5$

$s=(61)/5=12.2\ ounces$

dolphi86 [110]3 years ago

3 0

Answer:

Step-by-step explanation:

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3 years ago

Find the percent of soccer and cross country kids in the fall

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3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

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$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
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$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
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Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
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$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
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Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
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3 years ago

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zvonat [6]

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Step-by-step explanation:

LW(H/3)

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3 years ago

Read 2 more answers