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4 years ago

What is the real part of 4-5ia. 5b. 4c. -5id. -5PLEASE HELP!

Mathematics

1 answer:

marysya [2.9K]4 years ago

7 0

Answer:

Step-by-step explanation:

A complex number in its rectangular form is expressed as x+iy where x is the real part of the complex number and y is the imaginary because it is attached to the imaginary number i.

Given the complex number 4-5i, comparing the complex number with x+iy

4 = x and iy = -5i

Hence x = 4 and y = -5.

Since x is the real part of the complex number x+iy,<em> hence 4 will be the real part of the complex number 4-5i based on comparison. </em>

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The interior angles of a triangie have measures 100° (4y + 25) ° , and (3y + 20) What is the value of y?

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Step-by-step explanation:

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2 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

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Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

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