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3 years ago

One tank is filling at the rate of 3/4 gallon / 2/3 minute a second tank is filling at the rate of 3/8 gallon per 1/2 minute whi

ch tank is filling faster equation

Mathematics

1 answer:

eduard3 years ago

7 0

$\bf \cfrac{\quad \stackrel{gallons}{\frac{3}{4}}\quad }{\stackrel{minutes}{\frac{2}{3}}}\implies \cfrac{3}{4}\cdot \cfrac{3}{2}\implies \cfrac{9}{8}~g/m
\\\\\\
\cfrac{\quad \stackrel{gallons}{\frac{3}{8}}\quad }{\stackrel{minutes}{\frac{1}{2}}}\implies \cfrac{3}{8}\cdot \cfrac{2}{1}\implies \cfrac{3}{4}~g/m\\\\
-------------------------------$

$\bf \cfrac{3}{4}\cdot \cfrac{2}{2}\implies \cfrac{6}{8}\quad \textit{clearly then }\cfrac{9}{8}\textit{ is larger than }\cfrac{6}{8}
\\\\\\
\stackrel{\stackrel{tank's~volume}{gallons}}{y}=\cfrac{9}{8}\stackrel{minutes}{x}$

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Find the length of CD shown in red below. Show all work.

PIT_PIT [208]

By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.

<h3>How to calculate the length of an arc</h3>

The figure presents a circle, the arc of a circle (s), in inches, is equal to the product of the <em>central</em> angle (θ), in radians, and the radius (r), in inches. Please notice that a complete circle has a central angle of 360°.

If we know that θ = 52π/180 and r = 6 inches, then the length of the arc CD is:

s = [(360π/180) - (52π/180)] · (6 in)

s ≈ 32.254 in

By definition of circumference, the length of the arc EF (radius: 6 in, central angle: 308°) shown in red is approximately equal to 32.254 inches.

<h3>Remark</h3>

The statement has typing mistakes, correct form is shown below:

<em>Find the length of the arc EF shown in red below. Show all the work.</em>

To learn more on arcs: brainly.com/question/16765779

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2 years ago

Find the equation of the line in slope-intercept form. m=− 4 7 through (14, 3)

fomenos

I think your answer is y=-4/7x+5

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3 years ago

When adding vegetables to a footlong (30 cm) sandwich, how many ounces or slices of vegetables do you use? select all that apply

MA_775_DIABLO [31]

A footlong sandwich will fit other ingredients besides vegetables, such as 4 slices cheese or 4 strips of beacon, but if we only talk about vegetables, you can 1.5 oz. Lettuce and 1 oz. Onion, 6 slices each; Tomato, cucumber, pickles, pepper, olives, jalapeños.

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3 years ago

A batter hits a baseball 4 feet above the ground with an upward velocity of 59 feet per second. The function h(t) = -16t^2 + 49t

Anit [1.1K]

Answer:

3 seconds

Step-by-step explanation:

when h=7 ft

-16t²+49t+4=7

-16t²+49t-3=0

16t²-49t+3=0

16t^2-48t-t+3=0

16t(t-3)-1(t-3)=0

(t-3)(16t-1)=0

t=3,t=1/16

t=1/16 seconds is the time when it just started.

so reqd. time=3 seconds.

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3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

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

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

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

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

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3 years ago