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2 years ago

40 points!! What is the area of the quadrilateral with vertices at (2, -4), (4, -4), (2, 1), and (4, 1)?

Mathematics

1 answer:

stiv31 [10]2 years ago

8 0

Answer:Expand areaofthequadrilateralwithverticesat(2, - 4),(4, - 4),(2,1),and(4,1)

Step 1: Multiply (areaofthequadrilateralwithverticesat) * (2, - 4,)

Step 1.1: Multiply areaofthequadrilateralwithverticesat by each term in 2, - 4,

Step 1.1.1: Multiply areaofthequadrilateralwithverticesat * 2,

Group constants and powers by variable

(1 * 1)(a1 + 1 + 1 + 1 + 1 + 1r1 + 1 + 1 + 1e1 + 1 + 1 + 1 + 1o1f1t1 + 1 + 1 + 1 + 1h1 + 1q1u1d1i1 + 1 + 1l1 + 1w1v1c1s1)

a6r4e5oft5h2qudi3l2wvcs

Step 1.1.2: Multiply areaofthequadrilateralwithverticesat * -4,

Group constants and powers by variable

(1 * 1)(a1 + 1 + 1 + 1 + 1 + 1r1 + 1 + 1 + 1e1 + 1 + 1 + 1 + 1o1f1t1 + 1 + 1 + 1 + 1h1 + 1q1u1d1i1 + 1 + 1l1 + 1w1v1c1s1)

a6r4e5oft5h2qudi3l2wvcs

Step 1 answer = a6r4e5oft5h2qudi3l2wvcs + 1a6r4e5oft5h2qudi3l2wvcs

Group all like terms

2a6r4e5oft5h2qudi3l2wvcs

Step-by-step explanation:

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$