Help me please I need your help

Write a translation rule that maps point D(7, −3) onto point D'(2, 5).

12345 [234]

Answer:

The translation rule is described by $D'(x,y) =(x-5,y+8)$ .

Step-by-step explanation:

According to Linear Algebra, a translation consists in sum a given vector (original point in this case) with another vector (translation vector). We can define translation as follows:

$D'(x,y) = D(x,y) +U(x,y)$ (Eq. 1)

Where:

$D(x,y)$ - Original vector with respect to origin, dimensionless.

$D'(x,y)$ - Translated vector with respect to origin, dimensionless.

$U(x,y)$ - Translation vector with respect to original vector, dimensionless.

From (Eq. 1) we get that translation vector is:

$U(x,y) = D'(x,y)-D(x,y)$

If we know that $D(x,y) = (7,-3)$ and $D'(x,y) =(2,5)$ , then the translation vector is:

$U(x,y) = (2,5)-(7,-3)$

$U(x,y) = (-5,8)$

And we find the translation rule by assuming that $D(x,y) = (x,y)$ and $U(x,y) = (-5,8)$ in (Eq. 1):

$D'(x,y) = (x,y)+(-5,8)$

$D'(x,y) =(x-5,y+8)$

The translation rule is described by $D'(x,y) =(x-5,y+8)$ .

5 0

3 years ago

A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes.

masya89 [10]

Refer to the diagram below, which provides a plan view of the problem

Because each player applies the same force, F, the angle between the force and the direction of motion is (1/2)*25° = 12.5°.

Because the sled moves with constant velocity, the sled is in dynamic equilibrium, and
2F cos(12.5°) = 1140 N
1.9526 F = 1140
F = 583.84 N

Answer: 583.8 N (nearest tenth)

4 0

3 years ago

Will give brainliest.

Anon25 [30]

Answer:

4 days is the correct answer

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

(7+8)+(5+9)+4(2+5) =

Inessa [10]

Answer: 57

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago