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3 years ago

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Which of the following are equivalent to 104? Select all that apply.

1 answer:

solniwko [45]3 years ago

8 0

104 is the answer to your problem

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Whats the slope of the 2 points (6,-6) & (-1,-3)

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$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{-1}-\underset{x_1}{6}}}\implies \cfrac{-3+6}{-7}\implies -\cfrac{3}{7}$

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PLEASE HELP!!!<br> 9/10 divided by (-3/5) <br> Write the answer as a mixed number

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3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

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