The answer is √3-√2+<span>√6 = in decimal form 2.76732698</span><span /><span>
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<span>117.86 hope i helped!</span>
62=2w+2l
120= 2(2w)+ 2(l+21)
120=4w+2l+42
78= 4w+2l
39=2w+l
39-2w=l
Use substitution
62=2w+2(39-2w)
62= 2w+78-4w
62= -2w+78
-16=-2w
8=w
Plug w in
62=2(8)+2l
62=16+2l
46=2l
23=l
Final answer: Width=8, Length=23
Answer:
The point lie outside the circle
Step-by-step explanation:
step 1
Find the distance between the center of the circle and the given point
the formula to calculate the distance between two points is equal to
![d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y2-y1%29%5E%7B2%7D%2B%28x2-x1%29%5E%7B2%7D%7D)
we have
(0,0) and (-3,-7)
substitute the values
![d=\sqrt{(-7-0)^{2}+(-3-0)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-7-0%29%5E%7B2%7D%2B%28-3-0%29%5E%7B2%7D%7D)
![d=\sqrt{58}\ units](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B58%7D%5C%20units)
step 2
Compare the distance with the radius
If d=r ----> the point lie on the circle
If d > r -----> the point lie outside the circle
If d < r -----> the point lie inside the circle
we have
![d=\sqrt{58}\ units](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B58%7D%5C%20units)
![r=\sqrt{53}\ units](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B53%7D%5C%20units)
so
![\sqrt{58}\ units > \sqrt{53}\ units](https://tex.z-dn.net/?f=%5Csqrt%7B58%7D%5C%20units%20%3E%20%5Csqrt%7B53%7D%5C%20units)
![d > r](https://tex.z-dn.net/?f=d%20%3E%20r)
therefore
The point lie outside the circle