All categories

3 years ago

HELP high school pre cal

Mathematics

1 answer:

kogti [31]3 years ago

6 0

Answer:

The option A) -4-3i is correct

Therefore the simplified given expression is (-1-6i)+(-3+3i)=-4-3i

Step-by-step explanation:

Given that the expression is (-1-6i)+(-3+3i)

To simplify the given expression as below :

(-1-6i)+(-3+3i)=-1-6i-3+3i

=(-1-3)+(-6i+3i) ( combining the real parts and imaginary parts separately and adding the like terms )

=-4+(-3i)

=-4-3i

(-1-6i)+(-3+3i)=-4-3i

Therefore the simplified given expression is (-1-6i)+(-3+3i)=-4-3i

Therefore the option A) -4-3i is correct

You might be interested in

Find the equation of a straight line R, which passes through (50, 2) and is parallel to the line y + 10 = 2

Snowcat [4.5K]

Answer:

hii friends are you also follow me and give me brainliest

and so sorry I don't no answer

but you follow me and give me brainliest ok by

6 0

3 years ago

Imagine an experiment having three conditions and 20 subjects within each condition. The mean and variances of each condition ar

xxTIMURxx [149]

Answer:

1. Mean square B= 5.32

2. Mean square E= 16.067

3. F= 0.33

4. p-value: 0.28

Step-by-step explanation:

Hello!

You have the information of 3 groups of people.

Group 1

n₁= 20

X[bar]₁= 3.2

S₁²= 14.3

Group 2

n₂= 20

X[bar]₂= 4.2

S₂²= 17.2

Group 3

n₃= 20

X[bar]₃= 7.6

S₃²= 16.7

1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.

Df B= k-1 = 3-1= 2

Sum Square B is:

∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= general mean is the mean that results of all the groups together.

General mean:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5

Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64

Mean square B= Sum Square B/Df B= 10.64/2= 5.32

2. The mean square error (MSE) is the estimation of the variance error (σ $_{e}^2$ → $S_{e} ^{2}$ ), you have to use the following formula:

Se²= (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²

n₁+n₂+n₃-k

Se²= 19*14.3 + 19*17.2 + 19*16.7 = 915.8 = 16.067

20+20+20-3 57

DfE= N-k = 60-3= 57

3. To calculate the value of the statistic you have to divide the MSB by MSE

$F= \frac{Mean square B}{Mean square E} = \frac{5.32}{16.067} = 0.33$

4. P(F $_{2; 57}$ ≤ F) = P(F $_{2; 57}$ ≤ 0.33) = 0.28

I hope you have a SUPER day!

3 0

3 years ago

The curve

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$

y-coordinate

Substitute in x [Function]: $\displaystyle y = \sqrt{4 - 3}$
[√Radical] Subtract: $\displaystyle y = \sqrt{1}$
[√Radical] Evaluate: $\displaystyle y = 1$

∴ Coordinates of Point N is (4, 1).

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

4 0

3 years ago

Need help ASAP!!!!!!!

natima [27]

Answer:

Step-by-step explanation:

You can subtract normally when the square roots are the same( like in your problem) but the squares stay they same and the numbers on the outside change.

4 0

3 years ago

Read 2 more answers

I need help? Can you please tell me

BaLLatris [955]

Answer:

whats the questun

Step-by-step explanation:

7 0

2 years ago