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1 year ago

Factor completely 18x2 − 21x −15. 3(2x 1)(3x − 5) 3(2x − 5)(3x 1) 3(2x − 1)(3x 5) 3(6x 1)(x − 5).

Mathematics

1 answer:

sergeinik [125]1 year ago

3 0

The factor of the $18x^{2} -21x-15$ will be 3(3x-5)(2x+1).

<h3>What will be the factor of $18x^{2} -21x-15$ ?</h3>

Given quadratic equation is $18x^{2} -21x-15$

By taking 3 common from whole of the equation it becomes.

$3(6x^{2} -7x-5)$

now by factorization equation will be.

$3(6x^{2} -10x+3x-5)$

3[(2x(3x-5)+1(3x-5)]

Taking (3x-5) common from whole equation it will become

3(3x-5)(2x+1).

Hence the factor of the $18x^{2} -21x-15$ will be 3(3x-5)(2x+1).

To know more about factorization follow.

brainly.com/question/25829061

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lutik1710 [3]

Answer: each guest gets 4.19 cm²

Step-by-step explanation:

The cake is a circle. The area which he cuts for himself is a sector with a central angle of 120 degrees. The formula for determining the area of a sector is expressed as

Area of sector = θ/360 × πr²

Where

θ represents the central angle

π is a constant whose value is 3.14

r represents the radius of the circle.

From the diagram,

r = 4cm

Therefore, area of the cake that he cut for himself is

120/360 × 3.14 × 4²

= 16.75 cm²

The total area of the cake is

3.14 × 4² = 50.24 cm²

Therefore, the rest of the cake is

50.24 - 16.75 = 33.49 cm²

The amount that each guest gets would be

33.49/8 = 4.19 cm²

8 0

2 years ago

An inverse variation function has a k value of 8. Which ordered pair is on the graph of the function?

Vinvika [58]

The points on the graph of the inverse variation are of the form:

(x, 8/x)

<h3>Which ordered pairs are on the graph of the function?</h3>

An inverse variation function is written as:

y = k/x.

Here we know that k = 8.

y = 8/x

Then the points (x, y) on the graph of the function are of the form:

(x, 8/x).

So evaluating in different values of x, we can get different points on the graph:

if x = 1, the point is (1, 8)
if x = 2, the point is (2, 4)
if x = 3, the point is (3, 8/3)
if x = 4, the point is (4, 2)

And so on.

If you want to learn more about inverse variations:

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7 0

1 year ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
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Step-by-step explanation:

We are given the following and are trying to find the second derivative at <em>x</em> = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$