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3 years ago

Point m is the midpoint of kL. M(1,-1) and L(8,-7. What are the coordinates of point k?

Mathematics

1 answer:

Phoenix [80]3 years ago

3 0

Answer: (-6, 5)

<u>Step-by-step explanation:</u>

Use the Midpoint Formula: $\dfrac{(x_k, y_k)+(x_L,y_L)}{2}=(x_m, y_m)$

Separate the x's and y's and solve them individually:

$\dfrac{x_k+x_L}{2}=x_m\qquad \qquad \qquad \qquad\dfrac{y_k+y_L}{2}=y_m\\\\\\\dfrac{x_k+8}{2}=1\qquad \qquad \qquad \qquad \qquad \dfrac{y_k-7}{2}=-1\\\\\\x_k+8=2\qquad \qquad \qquad \qquad \qquad y_k-7=-2\\\\\\x_k\quad =-6 \qquad \qquad \qquad \qquad \qquad y_k\qquad =5$

So, the k-coordinate is (-6, 5)

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<h3>Solution:</h3>

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Step-by-step explanation:

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Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

W need to find :

Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.

For no values of x, we get indetermined form (i.e $\frac{0}{0}$ ), Hence there is no holes

Vertical Asymptotes:

Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

or x=4, Hence at x = 4, f(x) has a vertical asymptote

X -intercept :

Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

or f(x) = 1 = y

hence at y =f(x) = 1, we get horizontal asymptote

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3 years ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

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