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Orlov [11]
3 years ago
14

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

Mathematics
1 answer:
solniwko [45]3 years ago
8 0
104 is the answer to your problem
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9s + 29<br> what is the value of s
olga2289 [7]

Answer:

wheres the rest of the problem?

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Whats the slope of the 2 points (6,-6) &amp; (-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 0
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.

4 0
3 years ago
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