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3 years ago

6 times the sum of a number and 5

Mathematics

2 answers:

Firdavs [7]3 years ago

5 0

Answer: 6 x (x+5)

Step-by-step explanation: Since it is six times a number AND five, you want to do the addition equation first. To show this you place parentheses around a variable (x) plus 5. You add six in front of this because it comes first in the word problem.

KatRina [158]3 years ago

4 0

Answer:

6(x+5)

Step-by-step explanation:

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The profit P (in thousands of dollars) for a company spending an amount s (in thousands of dollars on advertising is

sattari [20]

Answer:

The company should spend $40 to yield a maximum profit.

The point of diminishing returns is (40, 3600).

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

Coordinate Planes

Coordinates (x, y) → (s, P)

Functions

Function Notation

Terms/Coefficients

Factoring/Expanding

Quadratics

Algebra II

Coordinate Planes

Maximums/Minimums

Calculus

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
f’(x) = c·nxⁿ⁻¹

1st Derivative Test - tells us where on the function f(x) does it have a relative maximum or minimum

Critical Numbers

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle P = \frac{-1}{10}s^3 + 6s^2 + 400$

Step 2: Differentiate

[Function] Derivative Property [Addition/Subtraction]: $\displaystyle P' = \frac{dP}{ds} \bigg[ \frac{-1}{10}s^3 \bigg] + \frac{dP}{ds} [ 6s^2 ] + \frac{dP}{ds} [ 400 ]$
[Derivative] Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle P' = \frac{-1}{10} \frac{dP}{ds} \bigg[ s^3 \bigg] + 6 \frac{dP}{ds} [ s^2 ] + \frac{dP}{ds} [ 400 ]$
[Derivative] Basic Power Rule: $\displaystyle P' = \frac{-1}{10}(3s^2) + 6(2s)$
[Derivative] Simplify: $\displaystyle P' = -\frac{3s^2}{10} + 12s$

Step 3: 1st Derivative Test

[Derivative] Set up: $\displaystyle 0 = -\frac{3s^2}{10} + 12s$
[Derivative] Factor: $\displaystyle 0 = \frac{-3s(s - 40)}{10}$
[Multiplication Property of Equality] Isolate s terms: $\displaystyle 0 = -3s(s - 40)$
[Solve] Find quadratic roots: $\displaystyle s = 0, 40$

∴ s = 0, 40 are our critical numbers.

Step 4: Find Profit

[Function] Substitute in s = 0: $\displaystyle P(0) = \frac{-1}{10}(0)^3 + 6(0)^2 + 400$
[Order of Operations] Evaluate: $\displaystyle P(0) = 400$
[Function] Substitute in s = 40: $\displaystyle P(40) = \frac{-1}{10}(40)^3 + 6(40)^2 + 400$
[Order of Operations] Evaluate: $\displaystyle P(40) = 3600$

We see that we will have a bigger profit when we spend s = $40.

∴ The maximum profit is $3600.

∴ The point of diminishing returns is ($40, $3600).

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

Unit: Differentiation (Applications)

5 0

2 years ago