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AleksandrR [38]
3 years ago
9

NEED HELP 10-14!! SERIOUS ANSWERS ONLY 27 POINTS!! THANKS!

Mathematics
1 answer:
Sergio039 [100]3 years ago
5 0

Answers:

10.

B: Austin did not combine -3 and -9 correctly.

11.

x=8.5

12.

2r-4s

13.

10x+3y

14.

k=2

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draw a perpendicular line from the directrix passing through the focus, this will be the line of symmetry. The vertex(h, k) will be located on the line half way between the focus and directrix. The distance from the focus to the vertex is called the focal length, call it a. The then equation is (x - h)^2 = 4a(y - k) the equation can be manipulated to y = 1/4a(x - h)^2 + k hope it helps

8 0
3 years ago
Read 2 more answers
a) Suppose f"(z) exists on an interval I and f(s) has a zero at three distinct points a < b< c on I. Show there is a pbint
Fynjy0 [20]

Answer:

We can find a root in the average point (a+b+c)/3

Step-By-Step Explanation:

We can use the Rolle Theorem. Since f is two times deribable on both [a,b] and [b,c], then there exists points x in (a,b) and y in (b,c) such that f'(x) = f'(y) = 0. Now, again by using Rolle Theorem, since f' is derivable in [x,y] (because it is a closed interval in I), then there exists s in [x,y] such that f''(s) = 0. This proves a.

The cube (x-a)(x-b)(x-c) has 3 roots, a, b and c and it is two times derivable because it is a polynomial. Hence we can use (a) to ensure that there is a root on I. Nevertheless, we can try to find the root manually:

f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)

f''(x) = (x-c)+(x-b)+(x-c)+(x-a)+(x-b)+(x-a) = 2(x-a)+2(x-b)+2(x-c) = 2( (x-a) + (x-b) + (x-c) ) = 6x - 2 (a+b+c).

We want x such that

6x - 2(a+b+c) = 0, or, equivalently,

3x - (a+b+c) = 0

Hence

x = (a+b+c)/3

Is a root of f'' in I.

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3 years ago
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2006053?

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1,440 is the answer i learned lcm last friday
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professor190 [17]

Answer:

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