If f(x) = 7x- 6, which of the following is the inverse of kx)?

Hi, please what's 4/5 - (-1/3) quick! I dont have that much time PLEASE

ipn [44]

Answer:

17/15

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the perimeter of △ABC?

timofeeve [1]

<h3>Given</h3>

ΔABC
A(-3, -1), B(0, 3), C(1, 2)

the length of the perimeter of ΔABC to the nearest tenth

<h3>Solution</h3>

The perimeter of a triangle is the sum of the lengths of its sides. The length of each side can be found using the Pythagorean theorem. Effectively, each pair of points is treated as the end-points of the hypotenuse of a right triangle with legs parallel to the x- and y-axes. The leg lengths are the differences betweeen the x- and y- coordinates of the points.

The difference of the x-coordinates of segment AB are 0-(-3) = 3. The y-coordinate difference is 3-(-1) = 4. So, the leg lengths of the right triangle whose hypotenuse is segment AB are 3 and 4. The Pythagorean theorem tells us

... AB² = 3² +4² = 9 +16 = 25

... AB = √25 = 5

You may recognize this as the 3-4-5 triangle often introduced as one of the first ones you play with when you learn the Pythagorean theorem.

LIkewise, segment AC has coordinate differences of ...

... C - A = (1, 2) -(-3, -1) = (4, 3)

These are the same leg lengths (in the other order) as for segment AB, so its length is also 5.

Segment BC has coordinate differences ...

... C - B = (1, 2) -(0, 3) = (1, -1)

The length of the line segment is figured as the root of the sum of squares, even though one of the coordinate differences is negative. The leg lengths of the right triangle used for finding the length of BC are the absolute value of these differences, or 1 and 1. Then the length BC is

... BC = √(1² +1²) = √2 ≈ 1.4

So the perimeter of the triangle ABC is

... AB + BC + AC = 5 + 1.4 + 5 = 11.4 . . . . perimeter of ∆ABC in units

_____

Please be aware that the advice to "round each step" is bad advice, in general. For real-world math problems, you only round the final result. You always carry at least enough precision in the numbers to ensure that there will not be any error in the final rounding.

In this problem, the only number that is not an integer is √2, so it doesn't really matter.

7 0

3 years ago

Five points are plotted on a cm grid.The 5 points are 5 vertices of a hexagon. Each side of the hexagon has the same length.Work

Eddi Din [679]

Answer:

(13, 5)

Step-by-step explanation:

6 0

2 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: $\displaystyle x = 4$