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4 years ago

Please help! 50 points!

Mathematics

2 answers:

8090 [49]4 years ago

4 0

Answer:

If we ignore the fact that Function 2 has a negative slope, then Function 2 has the greater rate of change.

Step-by-step explanation:

Function 1 has the rate of change m = 2/5, because every increase of 2 in y occurs in conjunction with an increase in x of 5.

Function 2 is a decreasing function (that is, its slope is negative). As we move from (0, 3) to (4.5, 0), x increases by 4.5 and y decreases by 3. Hence, the slope of Function 2 is m = rise / run = -3/4.5, or m = -2/3.

The absolute values of the slopes of the two functions are 2/5 and 2/3 respectively. If we ignore the fact that Function 2 has a negative slope, then Function 2 has the greater rate of change.

But if we take into account the signed number -2/3, then Function 1 has the greater slope because it's positive whereas that of Function 2 is negative.

My name is Ann [436]4 years ago

4 0

Answer:

If we ignore the fact that Function 2 has a negative slope, then Function 2 has the greater rate of change.

Step-by-step explanation:

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kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

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

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

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

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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