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Alina [70]
3 years ago
8

What is the solution set of the following equation? -5x x 9 = -4x 12

Mathematics
1 answer:
Rama09 [41]3 years ago
8 0
The solution of this equation is 16/15

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How do you do the last two questions?
Mashutka [201]

Answer:

P = 6200 / (1 + 5.2e^(0.0013t))

increases the fastest

Step-by-step explanation:

dP/dt = 0.0013 P (1 − P/6200)

Separate the variables.

dP / [P (1 − P/6200)] = 0.0013 dt

Multiply the left side by 6200 / 6200.

6200 dP / [P (6200 − P)] = 0.0013 dt

Factor P from the denominator.

6200 dP / [P² (6200/P − 1)] = 0.0013 dt

(6200/P²) dP / (6200/P − 1) = 0.0013 dt

Integrate.

ln│6200/P − 1│= 0.0013t + C

Solve for P.

6200/P − 1 = Ce^(0.0013t)

6200/P = 1 + Ce^(0.0013t)

P = 6200 / (1 + Ce^(0.0013t))

At t = 0, P = 1000.

1000 = 6200 / (1 + C)

1 + C = 6.2

C = 5.2

P = 6200 / (1 + 5.2e^(0.0013t))

You need to change the exponent from negative to positive.

The inflection points are where the population increases the fastest.

5 0
3 years ago
A house was valued at $100,000 in the year 1987. The value appreciated to $165,000 by the year 2002.
iris [78.8K]

Answer:

  A) The annual multiplier was 1.0339; the annual increase was 0.0339 of the value.

  B) 3.39% per year

  C) $182,000

Step-by-step explanation:

A) Let's let t represent years since 1987. Then we can fill in the numbers and solve for r.

  165000 = 100000(1 +r)^15

  1.65^(1/15) = 1 +r . . . . . divide by 100,000; take the 15th root

  1.03394855265 -1 = r ≈ 0.0339

The value was multiplied by about 1.0339 each year.

__

B) The value increased by about 3.39% per year.

__

C) S = $100,000(1.03394855265)^18 ≈ $182,000

8 0
3 years ago
How to isolate scalars from a matrix
I am Lyosha [343]
The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.
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3 years ago
Y=6cosx^2 (derivative)
Ksenya-84 [330]
Dy/dx = d/dx 6cosx^2
 = 12cosx d/dx cosx
 = 12cosx (-sinx)
 = -12cosxsinx
8 0
4 years ago
It costs $3.45 to buy 5/8 lb of chopped walnuts. How much would it cost to purchase 7.5 lbs of walnuts? Explain or show your
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