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3 years ago

What is 6 more than 7 times a number is 41

Mathematics

1 answer:

son4ous [18]3 years ago

8 0

the problem would be set up as 6+(7*x)=41

first u take off the number 6 by subtracting it from itself to cancel it out then u do the same to the number 41

7x =41 now u need to cancel out the number 7 attatched to the x so u divikde it by itself and then do the same to the other side so 41 divided by 7 =5.9

so x =5.9

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What’s the answer???

docker41 [41]

Answer: B, D, and E

Step-by-step explanation:

f(x) = x² - 1

(0, 1): 1 = (0)² - 1 ⇒ 1 = -1 FALSE

(1, 0): 0 = (1)² - 1 ⇒ 0 = 0 TRUE

(3, 5): 5 = (3)² - 1 ⇒ 5 = 8 FALSE

(5, 24): 24 = (5)² - 1 ⇒ 24 = 24 TRUE

(-2, 3): 3 = (-2)² - 1 ⇒ 3 = 3 TRUE

(-4, -17): -17 = (-4)² - 1 ⇒ -17 = 15 FALSE

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3 years ago

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babymother [125]

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3 years ago

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The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

puteri [66]

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Area of a Rectangle: A = lw

l is length
w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

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

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
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[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Unit: Implicit Differentiation

Book: College Calculus 10e

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3 years ago

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Ivenika [448]

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Equation at the end of step

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One solution was found :

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3 years ago

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