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2 years ago

A line contains the points (82, −96) and (87, −86).

Mathematics

1 answer:

Alik [6]2 years ago

8 0

Answer:

<h2>The answer is 2</h2>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(82, −96) and (87, −86)

We have

$m = \frac{ - 86 - - 96}{87 - 82} = \frac{ - 86 + 96}{5} = \frac{10}{5} = 2 \\$

We have the final answer as

Hope this helps you

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The coordinate values of the vertices of △klm are integers. triangle klm has vertices located at (1, 1), (5, 4) and (5, 1). whic

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The rest of the queastion is :
 A) (-1, 1), (-1, 4), (2, 1)
B) (0, 0), (3, 4), (0, 5)
C) (-1, 1), (-4, 5), (-1, 5)
D) (0, 0), (-5, 0), (0, 4)
==============================

The distance between two points (x₁,y₁),(x₂,y₂) = d
$d = \sqrt{ (x2-x1)^{2} + (y2-y1)^{2} }$

we have K(1, 1), L(5, 4) and M(5, 1)
we will find the length between each two points.
KL = $d = \sqrt{ (5-1)^{2} + (4-1)^{2} }$ = 5
LM = $d = \sqrt{ (5-5)^{2} + (1-4)^{2} }$ = 3
KM = $d = \sqrt{ (5-1)^{2} + (1-1)^{2} }$ = 4

now check each group from the choices
the group that will give the same result will be the correct choice.
The correct choice is C
C) (-1, 1), (-4, 5), (-1, 5)

check the answer;
K'(-1, 1), L'(-4, 5), M'(-1, 5)

K'L' = $d = \sqrt{ (-4+1)^{2} + (5-1)^{2} }$ = 5
L'M' = $d = \sqrt{ (-1+4)^{2} + (5-5)^{2} }$ = 3
K'M' = $d = \sqrt{ (-1+1)^{2} + (5-1)^{2} }$ = 4

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2 years ago

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Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real n

e-lub [12.9K]

Answer:

See Below

Step-by-step explanation:

The function is a piecewise function defined as:

$N(t)=\left \{ {{25t+150} \ 0 \leq t \leq 6 \atop {\frac{200+80t}{2+0.05t} \ t \geq 8}} \right.$

We need to find the limit of the function as t goes to infinity. This means what is the max value of fish in the pond given times goes to infinity (on an on).

We will take the 2nd part of the equation since t falls into that range, t is infinity, which is definitely greater than 8.

$\lim_{t \to \infty} \frac{200+80t}{2+0.05t} \\\lim_{n \to \infty} \frac{80t}{0.05t}\\ \lim_{n \to \infty} \frac{80}{0.05} =1600$

This means the maximum number of fish at this pond is 1600, no matter how long it goes on.

A function is continuous at a point if we have the limit and the functional value at that point same.

Functional value at t = 8 is (we use 2nd part of equation):

$\frac{200+80t}{2+0.05t}\\\frac{200+80(8)}{2+0.05(8)}\\=350$

We do have a value and limit also goes to this as t approaches 8.

So, function is continuous at t = 8

We want to find is there a "time" when the number of fishes in the pond is 250, during t from 0 to 6. We plug in 250 into N(t) and try to find t. Make sure to use the 1st part of the piece-wise function. Shown below:

$N(t)=25t+150\\250=25t+150\\25t=250-150\\25t=100\\t=4$

The time is 4 years when the number of fishes in the pond is 250

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2 years ago

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"Then I will need an average quality score of 3.28 to meet 85.28 goal."

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The employee needs an average quality score of 3.28 to meet the goal of 85.28

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