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pickupchik [31]
3 years ago
13

The graph shows a translation from preimage triangle ABC to image triangle JKL. Which of the following represents the translatio

n?

Mathematics
2 answers:
yarga [219]3 years ago
4 0
X+7 Y+4 . I am 100 percent sure.

gavmur [86]3 years ago
4 0

Answer:

The correct option is D.

Step-by-step explanation:

From the given graph it is clear that the vertices of triangle ABC are A(-10,-9), B(15,-10), C(-16,5) and the vertices of triangle JKL are J(-3,-5), K(-22,-6), L(-9,9).

It is given that the graph shows a translation from preimage triangle ABC to image triangle JKL.

(x,y)\rightarrow (x+a,y+b)            .... (1)

We know that image of A(-10,-9) is J(-3,5).

A(-10,-9)\rightarrow J(-3,-5)         .... (2)

From (1) and (2), we get

x=-10,y=-9,x+a=-3,y+b=-5

x+a=-3

(-10)+a=-3

a=-3+10=7

y+b=5

-9+b=-5

b=-5+9=4

The value of a is 7 and the value of b is 4. So the required rule for translation is

(x,y)\rightarrow (x+7,y+4)

Therefore option D is correct.

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Step-by-step explanation:

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Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1
ladessa [460]

Answer:

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

Step-by-step explanation:

Given the initial value problem \frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits  \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi  = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi  - \frac{1}{2}

The solution to the initial value problem will be;

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

5 0
3 years ago
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