Select the correct answer.

Mathematics

1 answer:

alex41 [277]3 years ago

4 0

Answer:

Option B

Step-by-step explanation:

Average rate of change of a function 'f' in the interval x = a and x = b is given by,

Average rate of change of change = $\frac{f(b)-f(a)}{b-a}$

Given function is,

g(x) = $\frac{5}{x-1}+2$

We have to calculate the rate of change in the interval [-4, 3],

Average rate of change = $\frac{g(3)-g(-4)}{3-(-4)}$

g(3) = $\frac{5}{3-1}+1$

= 2.5 + 1

= 3.5

g(-4) = $\frac{5}{-4-1}+1$

= -1 + 1

= 0

Average rate of change = $\frac{3.5-0}{3+4}$

= $\frac{3.5}{7}$

= $\frac{1}{2}$

Therefore, Option B will be the answer.

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The length of line segment AK is 640 meters. ∆ABC, ∆CDF, and ∆FJK are similar, and 2AC = CF = 2FK. The first pillar is 20 meters

Aleksandr [31]

A to K = 640 meters

2AC = CF = 2FK

2AC = 2(160) = 320
CF = 320
2FK = 2(160) = 320

AC = 160
CF = 320
FK = <u>160</u>
AK 640

BG = 20 m ; Area = (160m*20m) / 2 = 3,200/2 = 1,600 m²

20:160 = x : 320
20*320 = 160x
6,400 = 160x
6,400/160 = x
40 = x

Area of CDF = (320m*40m) / 2 = 12,800 / 2 = 6,400 m²

8 0

4 years ago

What is the equation of the quadratic graph with a focus of (3,4) and a drirectrix of y =8

Pie

check the picture below. So the parabola looks more or less like so.

bearing in mind that the distance "p" is the same from the vertex to the directrix or the focus point, so the vertex is therefore half-way between those two fellows, now, from 3,4 up to y = 8, there are 4 units, and the vertex is half-way, thus is at y = 6, and x = 3, (3, 6) as you see there in the picture.

the parabola is vertical, meaning the squared variable is the "x", and is opening downwards, meaning the "p" distance is negative, so since "p" is 2 units, then p = -2.

$\bf \textit{parabola vertex form with focus point distance}
\\\\
4p(y- k)=(x- h)^2
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\begin{cases}
h=3\\
k=6\\
p=-2
\end{cases}\implies 4(-2)(y-6)=(x-3)^2\implies -8(y-6)=(x-3)^2
\\\\\\
y-6=\cfrac{(x-3)^2}{-8}\implies \blacktriangleright y=-\cfrac{1}{8}(x-3)^2+6 \blacktriangleleft$