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Ludmilka [50]
3 years ago
7

A dolphin is swimming to the surface of the ocean while at the same time a bird is flying towards the surface of the ocean. The

function fx) models the position, in feet, of the dolphin x seconds after it starts swimming towards the surface. The function models the dolphin's position for the first 5 seconds The function g (x) models the position, in feet, of the bird x seconds after the bird begins flying towards the surface. The function models the bird's position for the first 5 seconds. A partial table of values ​​for the two linear functions is shown. x f (x) g (x) 0.0-11 10 0.59 9 1.07 8 1.55 7 2.0 -3 6 2.5 -1 5 After how many seconds will the dolphin and bird be at the same position?

Mathematics
2 answers:
maksim [4K]3 years ago
5 0

They are at the same position at 0.5 seconds because with respect to the ocean's surface, they are both 9 feet away.

Hope this helps!!

Len [333]3 years ago
5 0

Answer:

The dolphin and bird be at the same position will be at the same position after 3.5 seconds

Step-by-step explanation:

The general equation of a line is h(x) = m*x + b, where m is the slope and b is the y-axis interception. We can get m and b from data as follows. To get m, we need 2 points (x1,y1) and (x2, y2) and the next formula m = (y2 - y1)/(x2 - x1).  

For f(x), using (0.5, -9) and (0, -11)

m = (-9 - (-11))/(0.5 - 0) = 4

For g(x), using (0.5, 9) and (0, 10)  

m = (9 - 10)/(0.5 - 0) = -2

To get b, we need to see f(x) and g(x) values at x = 0. So, for f(x), b = -11 and for g(x), b = 10. Then,  

f(x) = 4*x - 11

g(x) = -2*x + 10

The dolphin and bird be at the same position when f(x) be equal to g(x). Mathematically :

f(x) = g(x)

4*x - 11 = -2*x + 10

4*x + 2*x = 10 + 11

6*x = 21

x = 21/6

x = 3.5 seconds

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Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
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\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
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= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
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