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4 years ago

Can someone please help me with the last question?

Mathematics

1 answer:

morpeh [17]4 years ago

3 0

Answer:

1/2

Step-by-step explanation:

The question appears to be asking only for the slope. The given point seems to be irrelevant.

The slope of the perpendicular line is the opposite of the reciprocal of the slope of the given line. That slope is the coefficient of x in the line's equation: -2. The opposite reciprocal of -2 is ...

-1/(-2) = 1/2

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3 years ago

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4 years ago

Which set of ordered pairs does not represent a function?

weqwewe [10]

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{(0,0),(0,1),(1,2),(1,3)}

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3 years ago

Solve for x: 2x - 2 > 4x + 6

laiz [17]

Isolate the x. Subtract 4x from both sides, and add 2 to both sides

2x (-4x) - 2 (+2) > 4x (-4x) + 6 (+2)

2x - 4x > 6 + 2

Simplify. Combine like terms

-2x > 8

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-2x/-2 > 8/-2

x > 8/-2

x < -4

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7 0

4 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

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

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

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

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

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

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.

3 0

4 years ago