I need help is it A B C or D

Consider the graph of the sixth-degree polynomial function f.

vekshin1

Answer:

b = 1, c = -1 and d = 4

Step-by-step explanation:

To solve this question the rule of multiplicity of a polynomial is to be followed.

If the multiplicity of a polynomial is even at a point, graph of the polynomial will touch the x-axis.

If the multiplicity of the polynomial is odd, graph will cross the x-axis at that point.

From the graph of function 'f',

f(x) = (x - b)(x - c)²(x - d)³

Since, graph of the function 'f' crosses x-axis at x = 1 and x = 4, multiplicity will be odd and touches the x-axis at x = -1 multiplicity will be even.

So the function will be,

f(x) = (x - 1)[x - (-1)]²(x - 4)³

Therefore, b = 1, c = -1 and d = 4 will be the answer.

7 0

3 years ago

Read 2 more answers

A tennis club has 560 members.

enyata [817]

Answer:

see explanation

Step-by-step explanation:

sum the parts of the ratio, 5 + 6 + 3 = 14 parts

Divide number of members by 14 to find the value of one part of the ratio.

560 ÷ 14 = 40 ← value of 1 part of the ratio

(i)

number of women = 6 × 40 = 240

(ii)

number of children = 3 × 40 = 120

8 0

3 years ago

The monthly revenue of a certain company is given by R 820p-7p, where p is the price in dollars of the product the company manuf

lora16 [44]

The revenue will be 12,000 when the price of the product is $100

Given :

The monthly revenue of a certain company

$R=820p-7p^2$

We need to find out p when Revenue R is 12000

Lets replace R with 12000 and solve for p

$12000=820p-7p^2\\820p-7p^2=12000\\-7p^2+820p-12000=0$

Apply quadratic formula

$p=\frac{-820\pm \sqrt{820^2-4\left(-7\right)\left(-12000\right)}}{2\left(-7\right)}\\p=\frac{-820\pm \:580}{2\left(-7\right)}\\p=\frac{-820+580}{2\left(-7\right)},\:p=\frac{-820-580}{2\left(-7\right)}\\p=\frac{120}{7},\:p=100$

Given that price must be greater than 50

So the revenue will be 12000 when the price of the product is $100

Learn more : brainly.com/question/21503855

5 0

2 years ago

Find <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \frac{dy}{dx} " alt=" \frac{d

nataly862011 [7]

Answer:

$\displaystyle y' = 2x + 3\sqrt{x} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

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

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

Identify

$\displaystyle y = (x + \sqrt{x})^2$

Step 2: Differentiate

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