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4 years ago

T (2,10) is the midpoint of CD. The coordinates of D are (2,13). What are the coordinates of C?

Mathematics

1 answer:

Sati [7]4 years ago

7 0

Your answer should be D because 13-10=3 and 10-3=7 (2, 7)!

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HELP with these questions

zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)

5 0

4 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$