Leon's older brother is 4 3/4 feet tall. Leon is 1/3 foot shorter than his brother . How tall is leon?

Mathematics

2 answers:

liberstina [14]3 years ago

8 0

Answer:

4 3/4 - 1/3 = 4 5/12

Step-by-step explanation:

4 9/12 - 4/12 = 4 5/12

Mademuasel [1]3 years ago

3 0

Let

x------> height of Leon's older brother

y------> height of Leon

Convert mixed number to an improper fraction

4\frac{3}{4}\ ft=\frac{4*4+3}{4}= \frac{19}{4}\ ft443 ft=44∗4+3=419 ft

we know that

x=\frac{19}{4}\ ftx=419 ft -----> equation A

y=x-\frac{1}{3}y=x−31 -----> equation B

substitute equation A in equation B

y=\frac{19}{4}-\frac{1}{3}=\frac{53}{12}\ fty=419−31=1253 ft

Convert to mixed number

\frac{53}{12}=\frac{48}{12}+\frac{5}{12}=4\frac{5}{12}\ ft1253=1248+125=4125 ft

therefore

the answer is

4\frac{5}{12}\ ft4125 ft

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Answer and step-by-step explanation:

The polar form of a complex number $a+ib$ is the number $re^{i\theta}$ where $r = \sqrt{a^2+b^2}$ is called the modulus and $\theta = tan^-^1 (\frac ba)$ is called the argument. You can switch back and forth between the two forms by either remembering the definitions or by graphing the number on Gauss plane. The advantage of using polar form is that when you multiply, divide or raise complex numbers in polar form you just multiply modules and add arguments.

(a) let's first calculate moduli and arguments

$r_1 = \sqrt{(-2\sqrt3)^2+2^2}=\sqrt{12+4} = 4\\ \theta_1 = tan^-^1(\frac{2}{-2\sqrt3}) =-\pi/6\\r_2=\sqrt{1^2+1^2}=\sqrt2\\ \theta_2 = tan^-^1(\frac 11)= \pi/4$

now we can write the two numbers as

$z_1=4e^{-i\frac \pi6}; z_2=e^{i\frac\pi4}$

(b) As noted above, the argument of the product is the sum of the arguments of the two numbers:

$Arg(z_1\cdot z_2) = Arg(z_1)+Arg(z_2) = -\frac \pi6 + \frac \pi4 = \frac\pi{12}$

(c) Similarly, when raising a complex number to any power, you raise the modulus to that power, and then multiply the argument for that value.

$(z_1)^1^2=[4e^{-i\frac \pi6}]^1^2=4^1^2\cdot (e^{-i\frac \pi6})^1^2=2^2^4\cdot e^{-i(12)\frac\pi6}\\=2^2^4 e^{-i\cdot2\pi}=2^2^4$

Now, in the last step I've used the fact that $e^{i(2k\pi+x)} = e^i^x ; k\in \mathbb Z$ , or in other words, the complex exponential is periodic with $2\pi$ as a period, same as sine and cosine. You can further compute that power of two with the help of a calculator, it is around 16 million, or leave it as is.