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3 years ago

13

Suppose the equation for line A is given by the equation y=--- 2. Ifline A and line B are perpendicular and the point (-4,1) lie

s on line B, then write an equation of line B.

1 answer:

nignag [31]3 years ago

4 0

Answer:

Not sure sorry

Step-by-step explanation:

not sure

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Y=2x-4 <br> What is the five ordered pairs and completing the input/output table

kati45 [8]

Answer:

{(0, -4), (1, -2), (2, 0), (3, 2), (4, 4)}

<u>Input values (x)</u> = {0, 1, 2, 3, 4}

<u>Output values (y)</u> = {-4, -2, 0, 2, 4}

Step-by-step explanation:

Given the linear equation, y = 2x - 4, you could practically choose any input value (for x), to have an output value (for y).

If x = 0:

y = 2(0) - 4

y = 0 - 4

y = -4

First ordered pair: y-intercept = (0, -4)

If x = 1:

y = 2(1) - 4

y= 2 - 4

y = -2

Second ordered pair: (1, -2).

If x = 2:

y = 2(2) - 4

y= 4 - 4

y = 0

Third ordered pair: x-intercept = (2, 0).

If x = 3:

y = 2(3) - 4

y = 6 - 4

y = 2

Fourth ordered pair: (3, 2).

If x = 4:

y = 2(4) - 4

y = 8 - 4

y = 4

Fifth ordered pair: (4, 4).

Your input values (x) = {0, 1, 2, 3, 4}

Your output values (y) = {-4, -2, 0, 2, 4}

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Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

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