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Andre45 [30]
3 years ago
10

The midpoint of A (-4, 2) and B(8, 5) is

Mathematics
2 answers:
Norma-Jean [14]3 years ago
8 0
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -4}}\quad ,&{{ 2}})\quad 
%  (c,d)
&({{ 8}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 8}} -4}{2}\quad ,\quad \cfrac{{{ 5}} + {{ 2}}}{2} \right)
inna [77]3 years ago
3 0

Answer:  The midpoint of the points A(-4, 2) and B(8, 5) is M(2, 3.5).

Step-by-step explanation:  We are given to find the midpoint of the points A(-4, 2) and B(8, 5).

We know that

the co-ordinates of the midpoint of the points (a, b) and (c, d) are given by

M=\left(\dfrac{a+c}{2},\dfrac{b+d}{2}\right).

Therefore, the co-ordinates of the midpoint of the points (-4, 2) and B(8, 5) will be

M=\left(\dfrac{-4+8}{2},\dfrac{2+5}{2}\right)=\left(\dfrac{4}{2},\dfrac{7}{2}\right)=(2,3.5).

Thus, the midpoint of the points A(-4, 2) and B(8, 5) is M(2, 3.5).

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3

Step-by-step explanation:

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Plugging 10 for k in the expression gives us
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4 0
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The number of bacteria in a petri dish is 50,000 (on day 0) if the amount is doubling each day by when will the number of bacter
Drupady [299]

Answer:

  • f(x) = 50000*2^x
  • 3 days

Step-by-step explanation:

<u>Given:</u>

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  • Growth rate = 2 times a day

<u>Required equation:</u>

  • f(x) = 50000*2^x

<u>Solve for x:</u>

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5 0
2 years ago
Darren wanted to see how contagious yawning can be. To better understand this, he conducted a social experiment in which he yawn
denis23 [38]

Answer:

10.5 minutes

Step-by-step explanation:

Thinking about the problem

The modeling function is of the form P(t)=A⋅Bf(t), where B=4B=4B, equals, 4 and f(t)=\dfrac{t}{10.5}f(t)=

10.5

t

​

f, left parenthesis, t, right parenthesis, equals, start fraction, t, divided by, 10, point, 5, end fraction.

Note that each time f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111, the quantity is multiplied by B=4B=4B, equals, 4.

Therefore, we need to find the ttt-interval over which f(t)f(t)f, left parenthesis, t, right parenthesis increases by 111.

Hint #22 / 3

Finding the appropriate unit interval

fff is a linear function whose slope is \dfrac{1}{10.5}

10.5

1

​

start fraction, 1, divided by, 10, point, 5, end fraction.

This means that whenever ttt increases by \Delta tΔtdelta, t, f(t)f(t)f, left parenthesis, t, right parenthesis increases by \dfrac{\Delta t}{10.5}

10.5

Δt

​

start fraction, delta, t, divided by, 10, point, 5, end fraction.

Therefore, for f(t)f(t)f, left parenthesis, t, right parenthesis to increase by 111, we need \Delta t=10.5Δt=10.5delta, t, equals, 10, point, 5. In other words, the ttt-interval we are looking for is 10.510.510, point, 5 minutes.

Hint #33 / 3

Summary

The number of people who yawned quadruples every 10.510.510, point, 5 minutes.

7 0
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How do linear and exponential functions compare over time?
Murljashka [212]

Answer:

Exponential Functions

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