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Lunna [17]
3 years ago
14

0=9 (k-2/3)+33solve for k

Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0
Your answer would be -3
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A blue whale calf weighed 2725 kilograms at birth. A blue whale calf gains 90 kilograms of weight each day for the first 240 day
Ivanshal [37]

Answer:the weight after 225 days is

22885 kilograms

Step-by-step explanation:

The initial weight of the blue whale calf at birth is 2725 kilograms. blue whale calf gains 90 kilograms of weight each day for the first 240 days after its birth. The weight increases in arithmetic progression. This means that the first term of the sequence, a is 2725, the common difference, d is 90.

The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

n is the number of terms of the sequence.

a is the first term

d is the common difference

We want to determine its weight, T225 after 225 days after it’s birth. It means that n = 225

Therefore

T225 = 2725 + (225 - 1)90

T225 = 2725 + 224×90 = 2725 + 20160

T225 = 22885

8 0
3 years ago
PLEASE HELP !<br> I AM RUNNING OUT OF TIME
andrezito [222]

Answer:

149

Step-by-step explanation:

If you order the numbers from smallest to biggest and then find the middle number it is 149

6 0
3 years ago
Read 2 more answers
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 <em>t</em> 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 <em>t</em> = 0 the value of <em>P</em> 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 <em>k</em> 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
3 years ago
Paola had 13 pounds of coffee beans to divide among 6 people. How many pounds of coffee should each person receive?
Juliette [100K]

Answer:

2

Step-by-step explanation:

here is how to do it 13/6=2

6 0
3 years ago
Match the situation to its corresponding graph.
Anna007 [38]
The answer is Graph D because it drops then it goes back up
8 0
4 years ago
Read 2 more answers
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