Condense the expression log 2x + log(x² - 9) - log x +3

1 answer:

4 0

Here the my solutions,hope you understand

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Zina [86]

$3 \frac{1}{4} = \frac{13}{4}$
$2 \frac{1}{3} = \frac{7}{3}$
$\frac{39}{12} + \frac{28}{12} =$
$= \frac{67}{12} = 5 \frac{7}{12}$

3 0

2 years ago

Gre4nikov [31]

Answer:

Step-by-step explanation:

We khow that the equation of a circle is written this way :

(x-a)²+(y-b)²=r² where (x,y) are the coordinates of the cercle's points and (a,b) the coordinates of the cercle's center and r the radius .

Our task is to khow the values of a and b :

We khow that the center is lying on the line 3x+2y=16⇒ 2y=-3x+16⇒ y= $\frac{-3}{2}$ x+8
We khow that the points P and Q are two points in the cercle
Let Ω be the center of this cercle
we can notice that : PΩ AND QΩ are both equal to the radius ⇒ PΩ=QΩ= r
So let's write the expression of this distance using vectors KHOWING THAT Ω(a,b)
Vector PΩ(a-4,b-6) and Vector QΩ(a-8,b-2)
PΩ= $\sqrt{(a-4)^{2}+(b-6)^{2} }$ and QΩ= $\sqrt{(a-8)^{2}+(b-2)^{2} }$
Let's substitute a by x and b by y
PΩ=QΩ we substitute each distance by its expression
After simplyfying the expressions we get finally : -12+8x-8y=0
now we have -12x +8x-8y=0 and the line equation 3x+2y-16=0
these are simultanious equations so after solving them we get x=3.8 wich is approximatively 4 and y=2
we substitute a by 4 and y by 2 in PΩ to get the radius
we get r = $\sqrt{(4-4)^{2}+(2-6)^{2} }$ = 4
so r²= 16
then the equation is : (x-4)²+(y-2)²=16