23=x-9;x+14 I need help with this problem.

Can someone please answer this for me and with solution and "Checking"( to see if your answer is correct) please.

mamaluj [8]

Answer:

√9/√13=5

√10-3r=r

Step-by-step explanation:

this must be the answer

4 0

3 years ago

Need help with this question please!

nevsk [136]

I would say it is A because if you subtract p, the original price by $2.50, you would get d, the discounted price. Look at B u see that you're adding the discount which doesn't make sense. Looking at C, the discounted price of different prices can't always be the same. And finally, D, the discounted price is greater than the original. Also, if you subtract you would get different discounts.

6 0

3 years ago

Find the derivative of

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$