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4 years ago

Complete the table of ordered pairs for the given liners equation

Mathematics

1 answer:

Papessa [141]4 years ago

5 0

Answer:

Step-by-step explanation:

Given equation of the line is,

y = -x + 1

We will get the output values (y-values) from the given equation for different input values (x-values).

For x = 0,

y = 0 + 1

y = 1

For x = -1,

y = -1 + 1

y = 0

For x = -6,

y = -6 + 1

y = -5

So the table of the ordered pairs will be,

x 0 -1 -6

y 1 0 -5

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Need help with this problem

Naya [18.7K]

Answer:

Slope is 1.5

Step-by-step explanation:

When one line has a slope of M, a perendicular line has a slope of -1/M

in this case M = -2/3, so you put it into -1/(-2/3) which equals to 1.5

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3 years ago

What is the solution of the inequality |2x+3|≤7?

Allushta [10]

Assuming you want to solve for x the result can vary. The result can be shown in multiple forms. Hope this helps!

Inequality Form:
−5≤x≤2
Interval Notation:
[−5,2]

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3 years ago

HELPPPPPPOOOO SHOW WORKING ALSO THANKS

Akimi4 [234]

Answer:

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Step-by-step explanation: LOLLLLL POINTS FORLIFE

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2 years ago

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<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

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3 years ago

Which of the following does not represent an attribute of the function f(x) = [x]?

luda_lava [24]

Answer: The graph has a line of symmetry at y = 0.

Step-by-step explanation:

I suppose that this refers to the function:

f(x) = IxI

The parent absolute value function.

Remember how this works:

IxI = x if x ≥ 0

IxI = -x if x ≤ 0.

Let's analyze the sentences:

1) "The x- and y-intercepts are (0,0)"

This is true, when x = 0 we have:

y = f(0) = I0I = 0

So the point (0, 0) is the only x-intercept and the only y-intercept.

2) "Domain: (-∞,∞); Range: [0,∞)"

This is also true, we clearly do not have any restriction in the domain.

And when looking at the range, we can see that the possible values of y are always positive or zero.

3) "(0, 0) is the minimum point."

Now, as x increases in the positive range, also does the value of y.

And as x decreases in the negative range, as the absolute value changes the sign, we will also have an increase in the value of y.

Then we have that (0, 0) must be the minimum.

4) "The graph has a line of symmetry at y = 0."

This is false.

The real line of symmetry is at x = 0, y = 0 corresponds to the x-axis, and as the graph opens up, we can not have symmetry about any line parallel to the x-axis.

4 0

4 years ago

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