Step-by-step explanation:
x-3>7 =>
x>10
hope this helps
Answer:
r = √13
Step-by-step explanation:
Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms: x^2 + 6x + y^2 -2y = 3. Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."
We need to "complete the square" of x^2 + 6x. I'll just jump in and do it: Take half of the coefficient of the x term and square it; add, and then subtract, this square from x^2 + 6x: x^2 + 6x + 3^2 - 3^2. Then do the same for y^2 - 2y: y^2 - 2y + 1^2 - 1^2.
Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2. Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1. Making these replacements:
(x + 3)^2 - 9 + (y - 1)^2 -1 = 3. Move the constants -9 and -1 to the other side of the equation: (x + 3)^2 + (y - 1)^2 = 3 + 9 + 1 = 13
Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r: r^2 = 13, so that the radius is r = √13.
Answer:
It is porportional
Step-by-step explanation:
Its Porportional because it goes through zero.
Answer:
1. S(1) = 1; S(n) = S(n-1) +n^2
2. see attached
3. neither
Step-by-step explanation:
1. The first step shows 1 square, so the first part of the recursive definition is ...
S(1) = 1
Each successive step has n^2 squares added to the number in the previous step. So, that part of the recursive definition is ...
S(n) = S(n-1) +n^2
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2. See the attachment for a graph.
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3. The recursive relation for an arithmetic function is of the form ...
S(n) = S(n-1) +k . . . . . for k = some constant
The recursive relation for a geometric function is of the form ...
S(n) = k·S(n-1) . . . . . . for k = some constant
The above recursive relation is not in either of these forms, so it is neither geometric nor arithmetic.
