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klemol [59]
4 years ago
14

Is 0.062 equal to .062

Mathematics
2 answers:
vitfil [10]4 years ago
7 0
Yes they are equal numbers. You can write them either way because in “.062” you just skip and don’t write a “0” in front.
Alona [7]4 years ago
4 0
 yes. The zero to the left of the decimal point does not affect the overall value of the problem when removed, since it is zero.
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Circles with centers $(2,1)$ and $(8,9)$ have radii $1$ and $9,$ respectively. the equation of a common external tangent to the
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For the complete understanding of the solution of this question please check the two attached files too.

Let us begin.

There are two circles given to us. Let us call them C1 and C2. Now, we know that C1 has centre at (2,1) and has a radius, say, r_{1} of 1. We also know that C2 has centre at (8,9) and has a radius, say, r_{2} of 9.

We know from coordinate geometry that of the center (a,b) and the radius, r of a circle are given then, the equation of the circle is given by:

(x-a)^2+(y-b)^=r^2

Using this formula and applying it to both the circles, we will get the equations of the two circles to be:

C1:(x-2)^2+(y-1)^2=1^2

or (x-2)^2+(y-1)^2-1=0 (subtracting both sides by 1)

and

C2: (x-8)^2+(y-9)^2=9^2

or (x-8)^2+(y-9)^2-81=0 (subtracting both sides by 9^2 which is 81).

Now, we know that, at if the circles have a common external tangent such that m or in other words m is negative then we must be having the condition that the circles touch each other as is evident from the first diagram.

All we need to do now is to equate the two equations of the circles C1 and C2 and solve for y in terms of x.

That will give us:

(x-2)^2+(y-1)^2-1=(x-8)^2+(y-9)^2-81=0

Solving this we get:

4y=-3x+15

y=-\frac{3}{4}x +\frac{15}{4}

y=-0.75x+3.75

This is the required equation of the tangent line and thus, as we can see b=\frac{15}{4} or 3.75

8 0
3 years ago
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