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Ksju [112]
3 years ago
11

Simplify the expression 4^-2*2^6 over 4^2*2^2

Mathematics
1 answer:
artcher [175]3 years ago
6 0

4=2^2\\------------\\\\\dfrac{4^{-2}\cdot2^6}{4^2\cdot2^2}=\dfrac{(2^2)^{-2}\cdot2^6}{(2^2)^2\cdot2^2}=\dfrac{2^{-4}\cdot2^6}{2^4\cdot2^2}=\dfrac{2^{-4+6}}{2^4\cdot2^2}=\dfrac{2^2}{2^4\cdot2^2}=\dfrac{1}{2^4}=\dfrac{1}{16}\\\\---------------------\\\\Used:\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}

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determine the point of intersection of the following pair of lines ............2x-3y-4=-13 and 5x=-2y+25. 3x+2y-7=0 and 2x=12-5y
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A nice graphing calculator app makes these trivially simple. (See the first two attachments.) It is available for phones, tablets, and as a web page.

__

The usual methods of solving a system of equations involve <em>elimination</em> or <em>substitution</em>.

There is another method that is relatively easy to use. It is a variation of "Cramer's Rule" and is fully equivalent to <em>elimination</em>. It makes use of a formula applied to the equation coefficients. The pattern of coefficients in the formula, and the formula itself are shown in the third attachment. I like this when the coefficient numbers are "too messy" for elimination or substitution to be used easily. It makes use of the equations in standard form.

_____

1. In standard form, your equations are ...

  • 2x -3y = -9
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Then the solution is ...

  x=\dfrac{-3(25)-(2)(-9)}{-3(5)-(2)(2)}=\dfrac{-57}{-19}=3\\\\y=\dfrac{-9(5)-(25)(2)}{-19}=\dfrac{-95}{-19}=5\\\\(x,y)=(3,5)

__

2. In standard form, your equations are ...

  • 3x +2y = 7
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Then the solution is ...

  x=\dfrac{2(12)-5(7)}{2(2)-5(3)}=\dfrac{-11}{-11}=1\\\\y=\dfrac{7(2)-12(3)}{-11}=\dfrac{-22}{-11}=2\\\\(x,y)=(1,2)

_____

<em>Note on Cramer's Rule</em>

The equation you will see for Cramer's Rule applied to a system of 2 equations in 2 unknowns will have the terms in numerator and denominator swapped: ec-bf, for example, instead of bf-ec. This effectively multiplies both numerator and denominator by -1, so has no effect on the result.

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