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nadezda [96]
3 years ago
13

A circle has its center at (1, 4) and a radius of 2 units. Find the equation of the circle using the Pythagorean Theorem.

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
8 0
(x-1)^2+(y-4)=4 is the answer but the pyth theorem is for triangles
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matrenka [14]

A.

So for 1 hour you pay 10 dolars then for 3 you pay 30 then for 11 you pay 110 then for 20 you pay 200.

For B and C I dont Know sorry.

4 0
3 years ago
What is the solution to the system of equations? Use the substitution method.
BabaBlast [244]
The answer is:
4x+2y=11}  4x+2y=11} x=2-2y
x-2=-2y    }  x+2y=2    } x=2-2(-3/6)
                                      x=2-(-6/6)
4(2-2y)+2y=11              x=2-(-1)
8-8y+2y=11                   x=3
8-6y=11
-6y=11-8
-6y=3/-6
y=-3/6=-1/2

So for my opinion the solution is A.(3,-1/2)
4 0
3 years ago
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
Which of the quadratic functions has the narrowest graph?
nadezda [96]
Hilf mir Baby, ich liebe dich
3 0
3 years ago
What is the solution to the following system of equations?<br><br> 3x - 2y = 4<br><br> x + 2y = 4
ivanzaharov [21]

Answer:

i have no clue lol

Step-by-step explanation:

i have no clue lol

4 0
3 years ago
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