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Blizzard [7]
1 year ago
5

A gallon of stain is enough to cover 200 square feet of decking. Bradley has two areas of decking he would like to cover with st

ain. One rectangular area is 23 feet by 10.4 feet, and the other is 10.5 feet by 7.2 feet. Which expression gives the number of gallons of stain Bradley will need?
Left-bracket (23) (10.5) + (10.4) (7.2) right-bracket divided by 200
Left-bracket (23) (10.4) + (10.5) (7.2) right-bracket divided by 200
Left-bracket (23) (10.5) + (10.4) (7.2) right-bracket times 200
Left-bracket (23) (10.4) + (10.5) (7.2) right-bracket times 200
Mathematics
1 answer:
Pani-rosa [81]1 year ago
7 0

The expression that we need to get is:

N = \frac{(23)*(10.4) + (10.5)*(7.2)}{200}

So the correct option is the second one.

<h3>Which expression gives the number of gallons of stain Bradley will need?</h3>

We know that 1 gallon is enough to cover 200 ft².

We have two rectangular areas, one of:

23 feet by 10.4 feet, and other of 10.5 feet by 7.2 feet.

Then the total area is:

A = (23 ft)*(10.4 ft) + (10.5ft)*(7.2 ft)

The number of gallons needed is given by the quotient between the area that we want to cover, and the area that covers one gallon, so the expression is:

N = \frac{(23ft)*(10.4ft) + (10.5ft)*(7.2ft)}{200ft^2} \\\\N = \frac{(23)*(10.4) + (10.5)*(7.2)}{200}

So the corerect option is the second one.

If you want to learn more about algebraic expressions:

brainly.com/question/4344214

#SPJ1

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A = 1.5 π in² ≈ 4.7 in²

Step-by-step explanation:

the area (A) of the sector is calculated as

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38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

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