-3*6 Multiplying and dividing integers

Mathematics

1 answer:

Vilka [71]2 years ago

7 0

All we need to do is multiply.

-3 * 6 = -18

Anytime you multiply a negative number to a positive number, its always going to be negative.

Best of Luck!

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Mei does 6 problems in 18 minutes. How many can she complete in 12 minutes

Dmitriy789 [7]

Do cross multiplication

6 = x
18 12

Now cross multiply

(6)(12) = (18)(x)

72 = 18x

Divide both sides by 18

72 = 18x
18 18

4 = x

So tMei can complete 4 problems in 12 minutes.

5 0

2 years ago

Find n(A) for A={0,1,2,3,...,2000}

miskamm [114]

There are 2001 items in that set.

So n(A) = 2001

We can individually count them all out, which is very slow and tedious, or we can use the formula
n-m+1

where,
m = starting value
n = ending value

In this case,
m = 0
n = 2000
So,
n-m+1 = 2000-0+1 = 2001

This formula only works if we increase the number by 1 each time (eg: 6, 7, 8, etc)

7 0

2 years ago

20 points and brainliest I’m in quiz in need it asap Number 4

iren [92.7K]

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.