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2 years ago

How can I complete this assignment.

Mathematics

1 answer:

Mademuasel [1]2 years ago

5 0

Answer:

what assignment.

Step-by-step explanation:

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Y=109x y=79x+2506 can some one solve both equations and a graph would be helpful plz

Amiraneli [1.4K]

Answer:

x ≈ 83.533

y = 9105.13

Step-by-step explanation:

Step 1: Substitution

109x = 79x + 2506

30x = 2506

x = 83.533

Step 2: Plug x in

y = 109(83.533)

y = 9105.13

Graphically:

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3 years ago

To change the 0.32 into percentage, I need to

Westkost [7]

Move the dot twice to the right

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2 years ago

For the following system, if you isolated x in the first equation to use the substitution method, what expression would you subs

Musya8 [376]

Answer:

x = 2y + 6

Step-by-step explanation:

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2 years ago

How to solve -3(x+4)=-15

egoroff_w [7]

Answer:

x=1 -3(1+4)=-15

Step-by-step explanation:

-3(x+4)=-15

---------------

-3 -3

x+4= 5

-4=-4

x=1

That should be right

3 0

3 years ago

Read 2 more answers

Find <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \frac{dy}{dx} " alt=" \frac{d

nataly862011 [7]

Answer:

$\displaystyle y' = 2x + 3\sqrt{x} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

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

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

Identify

$\displaystyle y = (x + \sqrt{x})^2$

Step 2: Differentiate

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

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

Unit: Derivatives

Book: College Calculus 10e

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2 years ago