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3 years ago

What is the domain of the function y= square root of x+6-7?

Mathematics

1 answer:

Rashid [163]3 years ago

8 0

Domain of y = sqrt(x + 6) - 7 
Finding domain is trying to check the following valid conditions 
 
condition 1: check for x such that the value in square root must greater or equal to 0 (applicable here) x + 6 >= 0 x > -6 ... this is the answer 
condition 2: check for x such that the divisor must not be zero (not applicable here) 
condition 3(misc) : check some functions for example ... log(0) is invalid...

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Does anyone know this ? i'm stuck.. i'll give you brainliest and 15 points !! pls show work.

insens350 [35]

Answer:

2y+3 is parallel or the same as QR and PS

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3 years ago

Geometry help please!

Gekata [30.6K]

given the following sets.

A = {0, 1, 2, 3}
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C = {0, a, 2, b}

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3 0

3 years ago

PLEASE HELP ASAP!!!!

Setler [38]

Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.

To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.

Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.

Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.

Therefore:

$\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 8} \\ f(4) = 3 \end{cases}}$

The nth root of a can contain negative number only if n is an odd number.

$\large{ \begin{cases} f( - 8 ) = \sqrt[3]{ - 2 \times - 2 \times - 2} \\ f(4) = 3 \end{cases}} \\ \large{ \begin{cases} f( - 8 ) = - 2\\ f(4) = 3 \end{cases}}$

Answer

f(-8) = -2
f(4) = 3

6 0

2 years ago

Help I cannot figure this question out.

netineya [11]

Answer:

B. x = -1 ± i

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

Standard Form: ax² + bx + c = 0
Factoring
Quadratic Formula: $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Algebra II

Imaginary Numbers: √-1 = i

Step-by-step explanation:

Step 1: Define

x² + 2x = -2

Step 2: Identify Variables

Rewrite Quadratic in Standard Form [Addition Property of Equality]: x² + 2x + 2 = 0
Break up Quadratic: a = 1, b = 2, c = 2

Step 3: Solve for x

Substitute in variables [Quadratic Formula]: $\displaystyle x=\frac{-2 \pm \sqrt{2^2-4(1)(2)}}{2(1)}$
[√Radical] Evaluate exponents: $\displaystyle x=\frac{-2 \pm \sqrt{4-4(1)(2)}}{2(1)}$
Multiply: $\displaystyle x=\frac{-2 \pm \sqrt{4-8}}{2}$
[√Radical] Subtract: $\displaystyle x=\frac{-2 \pm \sqrt{-4}}{2}$
[√Radical] Factor: $\displaystyle x=\frac{-2 \pm \sqrt{-1}\sqrt{4}}{2}$
[√Radicals] Simplify: $\displaystyle x=\frac{-2 \pm 2i}{2}$
Factor: $\displaystyle x=\frac{2(-1 \pm i)}{2}$
Divide: $\displaystyle x = -1 \pm i$

3 0

3 years ago

Help me out with the answer!!!!!!

Pie

Answer:

Adjacent

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers