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3 years ago

12

I need to know how to solve this equations

1 answer:

Vlad [161]3 years ago

3 0

The answer is x=12 hope it helps

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john goes out to eat at a local restaurant the bill comes to $47.38 john wants to leave a 20% tip how much of a tip should he le

Ghella [55]

Answer:

$56.86

Step-by-step explanation:

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3 years ago

In 2000 the population of a country reached 1 billion, and in 2025 it is projected to be 1.2 billion. (a) Find values for C an

Mice21 [21]

Answer:

(a) The value of C is 1.

(b) In 2010, the population would be 1.07555 billions.

(c) In 2047, the population would be 1.4 billions.

Step-by-step explanation:

(a) Here, the given function that shows the population(in billions) of the country in year x,

$P(x)=Ca^{x-2000}$

So, the population in 2000,

$P(2000)=Ca^{2000-2000}$

$=Ca^{0}$

$=C$

According to the question,

$P(2000)=1$

$\implies C=1$

(b) Similarly,

The population in 2025,

$P(2025)=Ca^{2025-2000}$

$=Ca^{25}$

$=a^{25}$ (∵ C = 1)

Again according to the question,

$P(2025)=1.2$

$a^{25}=1.2$

Taking ln both sides,

$\ln a^{25}=\ln 1.2$

$25\ln a = \ln 1.2$

$\ln a = \frac{\ln 1.2}{25}\approx 0.00729$

$a=e^{0.00729}=1.00731$

Thus, the function that shows the population in year x,

$P(x)=(1.00731)^{x-2000}$ ...... (1)

The population in 2010,

$P(2010)=(1.00731)^{2010-2000}=(1.00731)^{10}=1.07555$

Hence, the population in 2010 would be 1.07555 billions.

(c) If population P(x) = 1.4 billion,

Then, from equation (1),

$1.4=(1.00731)^{x-2000}$

$\ln 1.4=(x-2000)\ln 1.00731$

$0.33647 = (x-2000)0.00728$

$0.33647 = 0.00728x-14.56682$

$0.33647 + 14.56682 = 0.00728x$

$14.90329 = 0.00728x$

$\implies x=\frac{14.90329}{0.00728}\approx 2047$

Therefore, the country's population might reach 1.4 billion in 2047.

3 0

3 years ago

Please help me please

hram777 [196]

Answer:

detailed explanation is in image

Step-by-step explanation:

ans is 60 unit²

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X=1
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