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balandron [24]
3 years ago
8

X + .029 = .0078 .................

Mathematics
2 answers:
ioda3 years ago
6 0

Answer:

Omg i have that same question on a math test

Step-by-step explanation:

Nuetrik [128]3 years ago
4 0

Answer:

x=0.0212

Step-by-step explanation:

x=0.0212

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Deshawn invests $5,000 in a savings account that earns 6% annual interest, compounded continuously. How long will it take to dou
koban [17]

Answer:

Step-by-step explanation:

Investment = $5,000

Annual Interest = 5%

5000/100 x 5/1

50×5 = 250

First year interest = $250

Therefore, 250 x 20

= $5,000

It will take him 40 years.

6 0
4 years ago
Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

4 0
3 years ago
I Need help plz !!! just 2 questions Anyone !!!!
djverab [1.8K]
06.02 that he prediction is that he will land on heads and tails the same amount of times?   Maybe that helped you some
5 0
3 years ago
Read 2 more answers
7 L of an acid solution was mixed with 3 L of a 15% acid solution to make a 29% acid solution. What is the percent concentration
Setler [38]

well, let's say that the first solution is 7 liters and has "x" percent of acid, how much acid total in it? well, 7x, <u>keeping in mind that "x" is a decimal form</u>.

likewise, the second solution is 3 liters and has 15% acid, so how much acid is there in that one?  (15/100) * 3 = 0.45, so

\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&7&x&7x\\ \textit{2nd solution}&3&0.15&0.45\\ \cline{2-4}&\\ mixture&10&0.29&2.9 \end{array}~\hfill \implies 7x~~ + ~~0.45~~ = ~~2.9 \\\\[-0.35em] ~\dotfill\\\\ 7x=2.45\implies x=\cfrac{2.45}{7}\implies x=0.35~\hfill \stackrel{\textit{converting to \%}}{0.35\cdot 100}\implies \stackrel{\%}{35}

5 0
2 years ago
Read 2 more answers
5a-{3b-{4a-(5b-6a-7b)}]
saul85 [17]
5a-[3b-\{4a-(5b-6a-7b)\}]\\\\=5a-[3b-\{4a-(-6a-2b)\}]\\\\=5a-[3b-\{4a+6a+2b\}]\\\\=5a-[3b-\{10a+2b\}]\\\\=5a-[3b-10a-2b]\\\\=5a-[b-10a]\\\\=5a-b+10a\\\\\boxed{=15a-b}
4 0
3 years ago
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