(a) Let <em>x</em> = 3.12333…. Then 100<em>x</em> = 312.333… and 1000<em>x</em> = 3123.333…, so that
1000<em>x</em> - 100<em>x</em> = 3123.333… - 312.333…
==> 900<em>x</em> = 2811
==> <em>x</em> = 2811/900 = 937/300
(b) <em>x</em> = 4.38555… ==> 100<em>x</em> = 438.555… and 1000<em>x</em> = 4385.555…
==> 900<em>x</em> = 3947
==> <em>x</em> = 3947/900
(c) <em>x</em> = 37.222… ==> 10<em>x</em> = 372.222…
==> 9<em>x</em> = 335
==> <em>x</em> = 335/9
(d) <em>x</em> = 16.292929… ==> 100<em>x</em> = 1629.292929…
==> 99<em>x</em> = 1613
==> <em>x</em> = 1613/99
(e) <em>x</em> = 2.333… ==> 10<em>x</em> = 23.333…
==> 9<em>x</em> = 21
==> <em>x</em> = 21/9 = 7/3
Answer:
Step-by-step explanation:
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<u>Simplifying the expression using "a ÷ b = a x 1/b"</u>
X^2-6x+9
x^2-3x-3x+9
x(x-3)-3(x-3)
(x-3)(x-3)
(x-3)^2
To be a function every input value (x) can only have one output value (y)
On the graph there are 2 out put values for x = 2 so it is not a function.
The answer would be C.
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
