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2 years ago

please help me if you can if your going to say I'm cheating i don't care cause i know I'm not thanks!!

Mathematics

1 answer:

Dovator [93]2 years ago

4 0

Big fat badussy balls

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Please help

EleoNora [17]

Answer:

Step-by-step explanation:

(5⅔)/4 = (17/3)/4 = 17/12

7 0

3 years ago

What is the answer to this? i can’t figure it out.

Svetach [21]

9514 1404 393

Answer:

25 +0i

Step-by-step explanation:

The conjugate of a complex number is that number with the sign of the imaginary part reversed.

For z = -3+4i, its conjugate z* is -3-4i. The product of z and z* is ...

(-3 +4i)(-3 -4i) = -3(-3 -4i) +4i(-3 -4i)

= 9 +12i -12i -16i² = 9 +16 = 25

The real part of the product is 25; the imaginary part is 0.

(-3 +4i)(-3 -4i) = 25 +0i

_____

You may have noticed that (z)(z*) = |z|², the sum of the squares of the real and imaginary parts. It is always a non-negative real number.

8 0

2 years ago

What is the third quartile, Q3, of the following distribution?

GREYUIT [131]

Answer:

The third quartile is:

$Q_3=29$

Step-by-step explanation:

First organize the data from lowest to highest

4, 5, 10, 12, 14, 16, 18, 20, 21, 21, 22, 22, 24, 26, 29, 29, 33, 34, 43, 44

Notice that we have a quantity of n = 20 data

Use the following formula to calculate the third quartile $Q_3$

For a set of n data organized in the form:

$x_1, x_2, x_3, ..., x_n$

The third quartile is $Q_3$ :

$Q_3=x_{\frac{3}{4}(n+1)}$

With n=20

$Q_3=x_{\frac{3}{4}(20+1)}$

$Q_3=x_{15.75}$

The third quartile is between $x_{15}=29$ and $x_{16}=29$

Then

$Q_3 =x_{15} + 0.75*(x_{16}- x_{15})$

$Q_3 =29 + 0.75*(29- 29)\\\\Q_3 =29$

5 0

3 years ago

Read 2 more answers

My has 3 times more books that grant , and grant has 6 f

Klio2033 [76]

I’m not sure what happened while you were writing this, but if I’m reading it right, “My” has 18 books. If “My” has 3 times as many as 6, that also means 3 times 6, which is 18.

7 0

3 years ago

How do you solve sin (5pi/3) without a calculator?

Savatey [412]

very simple, we use the formula sin(a+b)=sinacosb + sinbcosa and sin(20)=2sinacosa

5pi = 2pi/3+3pi/3,

First, we use sin(a+b)=sinacosb + sinbcosa

sin(5pi/3)=sin(2pi/3+3pi/3)= sin(2pi/3+pi)= sin(2pi/3)cos(pi) +sin(pi)cos(2pi/3)

but we know that sin(pi)= 0, and cos (pi) = -1, so sin(5pi/3)= - sin(2pi/3)

now, use sin(2a)=2sinacosa, sin(5pi/3)= - sin(2pi/3)= -2sin(pi/3)cos(pi/3)

sin(5pi/3)= -2sin(pi/3)cos(pi/3)

sin(pi/3)= 0.86, cos(pi/3)=0.5, finally we have sin(5pi/3)= -0.86 x 0.5= -0.43

5 0

3 years ago

Read 2 more answers

please help me if you can if your going to say I'm cheating i don't care cause i know I'm not thanks!!​

please help me if you can if your going to say I'm cheating i don't care cause i know I'm not thanks!!