All categories

3 years ago

What is the slope of a line that passes through the origin and point (-3 -2) a. 2/3 b. -2/3 c. 3/2 or d. -3/2

2 answers:

8 0

The answer is C because if you invert it, the formula must work for the -3, -2 point and 0,0 (which is the origin point) another point would be 3,2 if you invert it

lions [1.4K]3 years ago

6 0

The origin is (0,0)

slope = (y2 - y1) / (x2 - x1)
(-3,-2)....x1 = -3 and y1 = -2
(0,0)...x2 = 0 and y2 = 0
now we sub
slope = (0 - (-2) / (0 - (-3) = (0 + 2) / (0 + 3) = 2/3 <=

You might be interested in

For every 10 flowers,there are 36 butterflies resting on the flower with the same number on each flower .How many butterflies ar

Gemiola [76]

Answer:

I don't know if this is right or not, but here's what I think.

There are 36 butterflies on each flower.

Step-by-step explanation:

Question:

For each 10 flowers, there are 36 butterflies resting on the flower with the same number on each flower. How many butterflies are on one flower?

3 0

3 years ago

The slope of the line whose equation is 3 y = 2 x - 3 is

fiasKO [112]

The photo has the answe

6 0

3 years ago

Please Help ASAP AP calculus Due in 1 day!!! Marking Brainliest!!!

MaRussiya [10]

Answer:

(D) $\displaystyle \frac{12}{1 + ln8}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integration

Integration Rule [Fundamental Theorem of Calculus 2]: $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt$

Step 2: Differentiate

Chain Rule: $\displaystyle f'(x) = \frac{d}{dx} \bigg[ \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt \bigg] \cdot \frac{d}{dx}[x^3]$
Integration Rule - Fundamental Theorem of Calculus 2: $\displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot \frac{d}{dx}[x^3]$
Basic Power Rule: $\displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot 3x^{3 - 1}$
Simplify: $\displaystyle f'(x) = \frac{3x^2}{1 + ln(x^3)}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle f'(2) = \frac{3(2)^2}{1 + ln(2^3)}$
Exponents: $\displaystyle f'(2) = \frac{3(4)}{1 + ln(8)}$
Multiply: $\displaystyle f'(2) = \frac{12}{1 + ln(8)}$