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3 years ago

What is the distance from one number to the next in a sequence of numbers that is

Mathematics

1 answer:

GrogVix [38]3 years ago

3 0

Answer:

common difference

Step-by-step explanation:

The consecutive terms in an arithmetic sequence have a common difference d between them.

If a is the first term in the sequence, then the terms are

a, a + d, a + 2d, a + 3d, ......... , a + (n - 1)d

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Answer:

$\frac{1}{16}$ $\frac{10}{9}$

Step-by-step explanation:

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16 * x = 1

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Amiraneli [1.4K]

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3 years ago

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":

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

8 0

3 years ago