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2 years ago

Write each points position as an ordered pair, like (1,2). A: B: C: D:

Mathematics

2 answers:

Charra [1.4K]2 years ago

6 0

Where is all the physics stuff

schepotkina [342]2 years ago

4 0

Answer:

A: (-7, 4)

B: (-6, -4)

C: (0, -3)

D: (2, 0)

Step-by-step explanation:

The first number in an ordered pair is the x point, so which number the point aligns with on the horizontal line.

The second number in an ordered pair is the y point, so which number the point aligns with on the vertical line.

If you look at point B you can see that it is directly below the -6 on the horizontal line and directly left of the -4 on the vertical line.

This means that point B is at (-6, 4)

Please give brainliest if I helped! :)

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hey, do any of you guys know how to plot functions on a graph. You know like f(X) and g(X)? I find it really hard and also do yo

Tpy6a [65]

It helps if you have an example, like f(x) = 2x+3
What you typically do, is:

- draw xy axis, label them (ie., 1,2,3,4 along both axes)
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- plot the calculated values as points. The calculated f(x) is your y value.
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3 years ago

A line has y intercept -6 and a x-intercept - 12. what is the equation of the line

koban [17]

Answer:

$\boxed{y = -\dfrac{1}{2}x-6}}$

Step-by-step explanation:

The equation for a straight line is

y = mx + b

where m is the slope of the line and b is the y-intercept.

The line passes through the points (-12, 0) and (0, 6)

(a) Calculate the slope of the line

$\begin{array}{rcl}m & =&\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\ & = &\dfrac{6 - 0 }{0 - (-12)}\\\\& = &\dfrac{6}{-12}\\\\& = & -\dfrac{1}{2}\\\end{array}$

(b) Write the equation

The y-intercept is at x = -6.

$\text{The equation for the line is $\boxed{\mathbf{y = -\dfrac{1}{2}x-6}}$}$

The diagram shows the graph of the line passing through the two intercepts.

5 0

3 years ago

If it is 2:00 what time would it be a half hour later

Ksju [112]

Answer:

2:30

Step-by-step explanation:

A half hour is 30 minutes.

7 0

3 years ago

Read 2 more answers

Between what two integers does -6.2 lie?

SCORPION-xisa [38]

Answer:

-6 and -7

Step-by-step explanation:

The best explanation is on a number line, -6.2 lies about 1/4 away from -6, and -6.2 lies .8 away from -7.

Best of Luck to you!

8 0

3 years ago

Read 2 more answers

Find the counterclockwise circulation and outward flux of the field F=7xyi+5y^2j around and over the boundary of the region C en

dezoksy [38]

Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).

Parameterize these pieces respectively by

r(t) = x(t) i + y(t) j = t i + t ² j

and

s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j

both with 0 ≤ t ≤ 1.

The circulation of F around ∂C is given by the line integral with respect to arc length,

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds$

where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :

• on the parabolic segment, we have

T = dr/dt = i + 2t j

• on the line segment,

T = ds/dt = -i - j

Then the circulation is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}$

Alternatively, we can use Green's theorem to compute the circulation, as

$\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}$

The flux of F across ∂C is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds$

where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.

• on the parabolic segment,

N = 2t i - j

• on the line segment,

N = - i + j

So the flux is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}$

5 0

3 years ago