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nirvana33 [79]
3 years ago
9

PLEASE ANSWERRRRRRRRRR

Mathematics
1 answer:
mash [69]3 years ago
8 0

Answer: OPTION B.

Step-by-step explanation:

Given the following System of equations:

\left \{ {{3x-5y=19} \atop {x+y=1}} \right.

You can use the Elimination Method to solve it. The steps are:

1. You can mutliply the second equation by -3.

2. Then you must add the equations.

3. Solve for the variable "y".

Then:

\left \{ {{3x-5y=19} \atop {-3x-3y=-3}} \right.\\.....................\\-8y=16\\\\y=\frac{16}{-8}\\\\y=-2

4. Now that you know the value of the variable "y", you must substitute it into any original equation.

5. The final step is to solve for "x" in order to find its value.

Then:

x+(-2)=1\\\\x=1+2\\\\x=3

Therefore, the solution is:

(3,-2)

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= [\frac{9}{2.6}  - \frac{2.5*2.5 }{2.5} ]^{2}

= [\frac{9}{2.6}  - \frac{2.5}{1} ]^{2}

*canceling 2.5 in numerator and denominator*

= [\frac{9-(2.5)(2.6)}{2.6} ]^2\\*Using L.C.M of 2.6 and 1 which comes out to be '2.6'= [\frac{9-(6.5)}{2.6} ]^2\\= [\frac{2.5}{2.6} ]^2\\*multiplying and dividing by '10'= [\frac{2.5*10}{2.6*10} ]^2\\= [\frac{25}{26} ]^2\\= \frac{25^2}{26^2}\\= \frac{625}{676}\\= 0.925

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Least Common Multiplier(LCM)

The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, LCM of 10, 15, and 20 is 60.

(b) [[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}    ] ^{2}

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[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}] ^{2}\\

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We can raise a power to a power

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This is called the power of a power property and says that to find a power of a power you just have to multiply the exponents.

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