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3 years ago

What is the product of (1 + i)(2 + i)?

Mathematics

2 answers:

Alex17521 [72]3 years ago

7 0

Answer:

The product of given expression is (1+3i).

Step-by-step explanation:

The given expression is

$(1+i)(2+i)$

Using distributive property of multiplication, we get

$1(2+i)+i(2+i)$

Using distributive property of multiplication again.

$2+i+i(2)+i(i)$

$2+i+2i+i^2$

$2+i+2i-1$ $[\because i^2=-1]$

Combine like terms.

$(2-1)+(i+2i)$

$1+3i$

Therefore the product of given expression is (1+3i).

Angelina_Jolie [31]3 years ago

3 0

(1+i)(2+i)

2+1i+2i-1

your answer: [1 + 3i]

hope this helps

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The formula Upper A equals 23.1 e Superscript 0.0152 tA=23.1e0.0152t models the population of a US state, A, in millions, t ye

Wewaii [24]

Answer:

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b) $t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years$

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

Step-by-step explanation:

For this case we assume the following model:

$A(t)= 23.1 e^{0.0152 t}$

Where t is the number of years after 2000/

Part a

For this case we want the population for 2000 and on this case the value of t=0 since we have 0 years after 2000. If we rpelace into the model we got:

$A(t=0)= 23.1 e^{0.0152(0)}=23.1e^0 =23.1$

So then the initial population at year 2000 is 23.1 million of people.

Part b

For this case we want to find the time t whn the population is 28.3 million.

So we need to solve this equation:

$28.3= 23.1 e^{0.0152(t)}$

We can divide both sides by 23.1 and we got:

$\frac{28.3}{23.1}= e^{0.0152t}$

Now we can apply natural log on both sides and we got:

$ln(\frac{28.3}{23.1})= 0.0152 t$

And then for t we got:

$t = \frac{ln(\frac{28.3}{23.1})}{0.0152}=13.357 years$

So for this case the answer would be 13.347 years, the population will be 28.3 million and ye year would be 2000+13.347 and that would be approximately in 2014

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