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3 years ago

Brainliest if answered correctly! Please help

Mathematics

1 answer:

sineoko [7]3 years ago

3 0

Go over 4 and up two on the negative side for (4-2) then just do the same for the other won but in the opposite square

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Elizabeth bought lemons (l) and apples (a). Each apple costs 35¢ and each lemon costs 15¢. All together, she bought 17 lemons an

Zina [86]

Step-by-step explanation:

Given:

Lemons - l and apples - a
a = 35¢, l = 15¢
Number of apples and lemons = 17
Cost of lemons and apples = $3.95 = 395¢

System as per question:

a + l = 17
35a + 15l = 395

a + l = 17
0.35a + 0.15l = 3.95

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3 years ago

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Choose the answer based on the most efficient method as presented in the lesson.

natita [175]

The next step is to add 9

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4 years ago

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X + y = 3 2x + 2y = -4

zaharov [31]

Answer:

(x,y)= (2/15, -(62/15)

Step-by-step explanation:

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3 years ago

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Find the quotient. (6x 2 - x - 40) ÷ (5 + 2x) 8 - 3x 3x + 8 3x - 8

solmaris [256]

6x^2 - x - 40 =(3x - 8)(2x + 5)
and
5 + 2x = 2x + 5

so
(6x^2 - x - 40)÷ (5 + 2x)
= [(3x - 8)(2x + 5)] ÷ (5 + 2x)
= 3x -8

answer
3x - 8

7 0

3 years ago

Suppose that the population P(t) of a country satisfies the differential equation dP/dt = kP (600 - P) with k constant. Its pop

jeka94

Answer:

The country's population for the year 2030 is 368.8 million.

Step-by-step explanation:

The differential equation is:

$\frac{dP}{dt}=kP(600 - P)\\\frac{dP}{P(600 - P)} =kdt$

Integrate the differential equation to determine the equation of P in terms of t as follows:

$\int\limits {\frac{1}{P(600-P)} } \, dP =k\int\limits {1} \, dt \\(\frac{1}{600} )[(\int\limits {\frac{1}{P} } \, dP) - (\int\limits {\frac{}{600-P} } \, dP)]=k\int\limits {1} \, dt\\\ln P-\ln (600-P)=600kt+C\\\ln (\frac{P}{600-P} )=600kt+C\\\frac{P}{600-P} = Ce^{600kt}$

At t = 0 the value of P is 300 million.

Determine the value of constant C as follows:

$\frac{P}{600-P} = Ce^{600kt}\\\frac{300}{600-300}=Ce^{600\times0\times k}\\\frac{1}{300} =C\times1\\C=\frac{1}{300}$

It is provided that the population growth rate is 1 million per year.

Then for the year 1961, the population is: P (1) = 301

Then $\frac{dP}{dt}=1$ .

Determine k as follows:

$\frac{dP}{dt}=kP(600 - P)\\1=k\times300(600-300)\\k=\frac{1}{90000}$

For the year 2030, P (2030) = P (70).

Determine the value of P (70) as follows:

$\frac{P(70)}{600-P(70)} = \frac{1}{300} e^{\frac{600\times 70}{90000}}\\\frac{P(70)}{600-P(70)} =1.595\\P(70)=957-1.595P(70)\\2.595P(70)=957\\P(70)=368.786$

Thus, the country's population for the year 2030 is 368.8 million.

3 0

4 years ago