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3 years ago

WILL GIVE BRAINLIEST!!!!!

Mathematics

2 answers:

Svetach [21]3 years ago

7 0

Answer:

1. 30x+60y=800

let 60y=800-30x

then:

30x+800-30x=800

0=0

2. You look nice and smart and a little mean

Darrick (DJ) Webb

Step-by-step explanation:

tatuchka [14]3 years ago

7 0

Answer:

infinitely many

Step-by-step explanation:

30x+60y=800

let 60y=800-30x

then:

30x+800-30x=800

0=0

infinitely many- unless you want to factor it instead of solving

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Tim has to read some books for school Tim read 1 books on Tuesday,

miv72 [106K]

Answer:

2 books

Step-by-step explanation:

1+1=2

7 0

3 years ago

Select the expression that is equivalent to the following: 12^-6 × 12^-5

Radda [10]

The solution to this problem is 743008370688

5 0

3 years ago

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

Find the missing term (x^12)^5*(x^-2)^9___=(x^40)^5

ratelena [41]

(x^12)^5=x^(12*5)=x^60
(x^(-2))^9=x^(-18)
x^60*x^(-18)=x^(40) and we need to find x^a such that
x^(42)*x^a=x^(40)^5
x^a=x^(40)^5*x^(-42)
x^a=x^(200)x^(-42)
x^a=x^(200-42)
x^a=x^158
so the missing term is x^158

6 0

3 years ago

15(x -20)-21= -3(x+1)+6

jeyben [28]

Answer:

x = 18

Step-by-step explanation:

Solve for x:

15 (x - 20) - 21 = 6 - 3 (x + 1)

Hint: | Distribute 15 over x - 20.

15 (x - 20) = 15 x - 300:

15 x - 300 - 21 = 6 - 3 (x + 1)

Hint: | Group like terms in 15 x - 300 - 21.

Grouping like terms, 15 x - 300 - 21 = 15 x + (-300 - 21):

15 x + (-300 - 21) = 6 - 3 (x + 1)

Hint: | Evaluate -300 - 21.

-300 - 21 = -321:

15 x + -321 = 6 - 3 (x + 1)

Hint: | Distribute -3 over x + 1.

-3 (x + 1) = -3 x - 3:

15 x - 321 = -3 x - 3 + 6

Hint: | Group like terms in -3 x - 3 + 6.

Grouping like terms, -3 x - 3 + 6 = (6 - 3) - 3 x:

15 x - 321 = (6 - 3) - 3 x

Hint: | Evaluate 6 - 3.

6 - 3 = 3:

15 x - 321 = 3 - 3 x

Hint: | Move terms with x to the left hand side.

Add 3 x to both sides:

15 x + 3 x - 321 = (3 x - 3 x) + 3

Hint: | Look for the difference of two identical terms.

3 x - 3 x = 0:

15 x + 3 x - 321 = 3

Hint: | Group like terms in 15 x + 3 x - 321.

Grouping like terms, 15 x + 3 x - 321 = (15 x + 3 x) - 321:

(15 x + 3 x) - 321 = 3

Hint: | Add like terms in 15 x + 3 x.

15 x + 3 x = 18 x:

18 x - 321 = 3

Hint: | Isolate terms with x to the left hand side.

Add 321 to both sides:

18 x + (321 - 321) = 321 + 3

Hint: | Look for the difference of two identical terms.

321 - 321 = 0:

18 x = 3 + 321

Hint: | Evaluate 3 + 321.

3 + 321 = 324:

18 x = 324

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 18 x = 324 by 18:

(18 x)/18 = 324/18

Hint: | Any nonzero number divided by itself is one.

18/18 = 1:

x = 324/18

Hint: | Reduce 324/18 to lowest terms. Start by finding the GCD of 324 and 18.

The gcd of 324 and 18 is 18, so 324/18 = (18×18)/(18×1) = 18/18×18 = 18:

Answer: x = 18

8 0

3 years ago