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3 years ago

I’ll give brainlest to the first one who answer the one my mouse is on it’s not that one!

Mathematics

1 answer:

Natalija [7]3 years ago

4 0

Answer:

The answer is A

Step-by-step explanation:

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It a math question so please help me out.

il63 [147K]

Answer: I say B

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Match the solution set given in inequality notation with the solution set given in interval motion x > 7.8

Maslowich

Answer:

(7.8, ∞)

Step-by-step explanation:

Given the inequality :

x > 7.8 ; the inequality can be explicitly interpreted as x greater than 8 ; this means that the inequality holds for all values of x above 7.8

The interval for which the inequality holds true is :

x > 7. 8 - - - > (7.8, ∞)

7 0

3 years ago

On a social studies test, James scored a 2025, Jules earned an 81%, and Scott scored a 0.815. Who earned the highest grade? Just

Airida [17]

Answer:

Scott

Step-by-step explanation:

20/25=.8 .8=80%

.815*100=81.5%

81%

After finding a decimal and multiplying by 100, you get the percent so you can compare. 81.5% is the highest, so Scott had the highest score.

Hope this helps. Please put brainliest if you can!

7 0

3 years ago

Which statements can be used to compare the characteristics of the functions

Aleks [24]

I think we need a picture to solve that

4 0

3 years ago

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

4 0

3 years ago