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ladessa [460]
3 years ago
14

Why the output is a zero number ?

D%200" id="TexFormula1" title="(x - 3) \times (x + 4) = 0" alt="(x - 3) \times (x + 4) = 0" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
ELEN [110]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

Given

(x - 3) × (x + 4) = 0, that is

(x - 3)(x + 4) = 0

The zero product indicates that (x - 3) = 0 or (x + 4) = 0

x - 3 = 0 ⇒ x = 3

x + 4 = 0 ⇒ x = - 4

Thus if x = 3, then

(3 - 3)(x + 4) = 0 × (x + 4) = 0

Similarly if x = - 4 the output is zero

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Write the equation of a line that includes the point (4, –2) and has a slope of 0 in slope-intercept form.
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Find the b value by plugging in the slope and known coordinates,

-2=0(4)+b
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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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2 years ago
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Answer:

Any-where below 1 and above 0

Step-by-step explanation:

I had this problem in math and my teacher told me this

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3 years ago
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