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TEA [102]
3 years ago
6

I’m confused on this one

Mathematics
1 answer:
andriy [413]3 years ago
6 0

The point-slope form of the equation of a line is

y - y_1 = m(x - x_1)

where (x1, y1) is a point on the line, and m is the slope of the line.

y - 3 = \dfrac{5}{3}(x - 0)

y - 3 = \dfrac{5}{3}x

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Use complete sentences to describe the transformation that maps ABCD onto its image.
Lapatulllka [165]

General Idea:

When a point or figure on a coordinate plane is moved by sliding it to the right or left or up or down, the movement is called a translation.

Say a point P(x, y) moves up or down ' k ' units, then we can represent that transformation by adding or subtracting respectively 'k' unit to the y-coordinate of the point P.

In the same way if P(x, y) moves right or left ' h ' units, then we can represent that transformation by adding or subtracting respectively 'h' units to the x-coordinate.

P(x, y) becomes P'(x\pm h, y\pm k). We need to use ' + ' sign for 'up' or 'right' translation and use ' - ' sign for ' down' or 'left' translation.

Applying the concept:

The point A of Pre-image is (0, 0). And the point A' of image after translation is (5, 2). We can notice that all the points from the pre-image moves 'UP' 2 units and 'RIGHT' 5 units.

Conclusion:

The transformation that maps ABCD onto its image is translation given by (x + 5, y + 2),

In other words, we can say ABCD is translated 5 units RIGHT and 2 units UP to get to A'B'C'D'.

6 0
3 years ago
I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

to

\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
3 years ago
Read 2 more answers
Can somebody plz use the diagram at the bottom to answer question 10 (only if u know how to do this) thx sm!
jonny [76]

Answer:

isosceles: 2 angles are the same and 2 sides are the same. Equilateral have all the same sides and angles. Saclene have all dfferent angles and sides.

11) isosceles

12) equilateral

13) isosceles or scalene unclear

---------{NOTICE}----------

If you have any additional questions the best way to reach me is via discord

user: michaelsaltandpepper#0001

5 0
3 years ago
Read 2 more answers
Angle D has a measure between 0° and 360° and is coterminal with a –920° angle. What is the measure of angle D?
topjm [15]
Take -920/360 = -2.55555
Take the fraction part -0.5555555 * 360 = -200
The positive of that would be 160 degrees
7 0
3 years ago
Read 2 more answers
Write an equation in​ point-slope form of the line that passes through the given​ points, then write the equation in​ slope-inte
GalinKa [24]
Y=2/9x+9

y-y1=m(x-x1) is the formula you use

5 0
3 years ago
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