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3 years ago

Can someone please explain to me how to add and subtract integers without using a number line? What's the rule for it?

1 answer:

4 0

ADD:
If both numbers are positive or negative then you have to add them if its a negative then your answer is a negative.6+1=7;-4+-1=-5

If one of the numbers are negative and the other one is positive then you will have to subtract.Then find the bigger number in the problem if its a negativie then your answer is a negative if your number is positive then your answer is a positive.-9+1=-8;4+-3=1

Sorry Don't Know What The Rules For Subtraction.But Hope This Helped You For Adding.

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Gemma and Leah are both jewelry makers. They made 39 necklaces. Each necklace has exactly 104 beads on it. How many beads did th

Kay [80]

Answer:

4056

Step-by-step explanation:

39 necklaces

104 beads/necklace

number of beads = number of necklaces * number of beads per necklace

number of beads = 39 * 104 = 4056

8 0

3 years ago

Find the slope of the line y= -8/5x + 2

VikaD [51]

Step-by-step explanation:

Given

y = - 8 /5 x + 2

Comparing the given equation with y = mx + c

slope (m) = - 8/5

Hope it will help :)

7 0

3 years ago

Winston and Sally are scuba diving. Winston is 27.77 meters below the surface of the ocean, and Sally is 2.10 meters deeper than

Nitella [24]

Answer:

-29.87

Step-by-step explanation:

8 0

3 years ago

The curve

kherson [118]

Answer:

Point N(4, 1)

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

Coordinates (x, y)
Functions
Function Notation
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

Derivative of a constant is 0

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

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$

y-coordinate

Substitute in x [Function]: $\displaystyle y = \sqrt{4 - 3}$
[√Radical] Subtract: $\displaystyle y = \sqrt{1}$
[√Radical] Evaluate: $\displaystyle y = 1$

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

4 0

2 years ago

Simplify

OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

$\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}$

we need to simplify that

$x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)$

so we can continue

$\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}$

and we can put all together

$\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}$

3 0

3 years ago