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3 years ago

Evaluate:-19 - (-10) + (20) - (30) - (-15) A) -4 B) -3 C) 3 D) 4

2 answers:

4 0

So firstly, remember that subtracting a negative number is the same as adding a positive number so our expression will look like this:

$-19+10+20-30+15$

Now solve as such:

$\underbrace{-19+10}_{-9}+20-30+15\\\\\underbrace{-9+20}_{11}-30+15\\\\\underbrace{11-30}_{-19}+15\\\\\underbrace{-19+15}_{-4}\\\\-4$

Your answer is -4, or A.

Ede4ka [16]3 years ago

4 0

Answer: A) -4

Step-by-step explanation: usatestprep approved

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Marysya12 [62]

Answer:

volume is just defined as multplying all the sides and is alot different from surface area, so it would be 1/5x1/5x/14 which is 1/100

Step-by-step explanation:

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3 years ago

Please help. the packet is due tonight

Zolol [24]

Answer:

[C] $\displaystyle \frac{-3}{250}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Factoring
Functions
Function Notation
Conjugations

Calculus

Limits
Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$
Limit Property [Multiplied Constant]: $\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$
Derivatives
Definition of a Derivative: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$\displaystyle f(x) = \frac{3}{\sqrt{x - 4}}$

$\displaystyle g(29)$

Step 2: Differentiate

Substitute in function [Function g(x)]: $\displaystyle g(x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h - 4}} - \frac{3}{\sqrt{x - 4}}}{h}$
Substitute in x [Function g(x)]: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{29 + h - 4}} - \frac{3}{\sqrt{29 - 4}}}{h}$
Simplify: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{25 + h}} - \frac{3}{5}}{h}$
Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15}{5\sqrt{25 + h}} - \frac{3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$
[Subtraction] Combine like terms: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15 - 3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$
Factor: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3(5 - \sqrt{25 + h})}{5\sqrt{25 + h}}}{h}$
Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{3(5 - \sqrt{25 + h})}{5h\sqrt{25 + h}}$
Rewrite [Limit Property - Multiplied Constant]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}}$
Root Conjugation: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}} \cdot \frac{5 + \sqrt{25 + h}}{5 + \sqrt{25 + h}}$
Multiply: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{5h\sqrt{25 + h} + h^2 + 25h}$
Factor: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{h(5\sqrt{25 + h} + h + 25)}$
Simplify: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + h} + h + 25}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + 0} + 0 + 25}$
Simplify: $\displaystyle g(29) = \frac{3}{5} \cdot \frac{-1}{50}$
Multiply: $\displaystyle g(29) = \frac{-3}{250}$

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

Unit: Derivatives

Book: College Calculus 10e

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3 years ago

Raymond just got done jumping at Super Bounce Trampoline Center. The total cost of his session was $43.25. He has to pay a $7 en

pantera1 [17]

7+1.25x=43.25
-7. - 7.00
____________
1.25x = 36.25
___________
1.25x. 1.25
x = 29

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3 years ago

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Did I do this right?

Dimas [21]

Yeah that’s correct

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3 years ago

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A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

nasty-shy [4]

Answer:

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Step-by-step explanation:

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6 0

3 years ago