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mojhsa [17]
2 years ago
8

PLEASE HELP DUE TODAY

Mathematics
1 answer:
AleksAgata [21]2 years ago
4 0

9514 1404 393

Answer:

  16 square units

Step-by-step explanation:

When you plot the points, you see they define a trapezoid with bases of lengths 2 and 6, and a height of 4. The area formula is ...

  A = (1/2)(b1 +b2)h

  A = (1/2)(2 +6)(4) = 16

The area of the trapezoid is 16 square units.

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If the sphere shown above has a radius of 10 units then what is the approximate volume of the sphere
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PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%
Yanka [14]

Answer:

x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}

Second equation:

\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}

Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}

Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12

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