If you were to exttend your number line to reach from 0 to 2 there would be five fifths for every whole number lenghts

Am I right ? Because I need help with this please thanks.

bazaltina [42]

to find the hypotenuse (or the ramp in question 4) you can use the equation

$a^{2} +b^{2} =c^{2}$

with a and b being the 3 feet and 9 feet and c being the unknown (ramp).

So if you plug 3 and 9 into the equation, it looks like this.

$3^{2} +9^{2} =c^{2}$

then square them.

$9+81=c^{2}$

simplify.

$90=c^{2}$

take the square root of both sides.

$\sqrt{90} = c$

$\sqrt{90}$ is approx. 9.5 so the answer would be H.

7 0

3 years ago

Corrine wants to purchase 9.6 pounds of trail mix. Each pound costs $1.20. How much will Corrine spend on trail mix?

Grace [21]

Multiply 9.6 pounds of trail mix by $1.20.

9.6 * 1.20 = 11.52

Corrine will spend $11.52 on trail mix.

7 0

3 years ago

Read 2 more answers

Please help me with this! show that {1,2} ⊆ {x|x^4 - x^3 - x^2 - 5x + 6 = 0}

Llana [10]

The set $\{x\mid x^4-x^3-x^2-6x+6=0\}$ is the set of all values of $x$ that are solutions to the given equation. So all you need to do is check whether $x=1$ and $x=2$ satisfy the equation.

$x=1\implies 1^4-1^3-1^2-5+6=0$

$x=2\implies2^4-2^3-2^2-10+6=0$

The equation holds, so both 1 and 2 are solutions, and hence $\{1,2\}$ is a subset of the solutions.

6 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

If 80 persons can perform a piece of work in 16 days of 10 hours each, how

eduard

Answer:

The number of men needed to perform a piece of work twice as great in tenth part of the time working 8 hours a day supposing that three of the second set can do as much work as four of the first set is:

1200 men.

Step-by-step explanation:

To find the answer, first, we're gonna find how many hours take to make the piece of work in 16 days, taking into account each day just has 10 hours:

Number of hours to make a piece of work = 16 * 10 hours
Number of hours to make a piece of work = 160 hours.

Now, we divide the total hours among the number of persons:

Equivalence of hours per person = 160 hours / 80 persons.
Equivalence of hours per person = 2 hours /person

This equivalence isn't the real work of each person, we only need this value to make the next calculations. Now, we have a piece of work twice as great as the first, then, we can calculate the hours the piece of work needs to perform it (twice!):

Number of hours to make the second piece of work = 160 hours * 2
Number of hours to make the second piece of work = 320 hours

We need to make this work in tenth part of the time working 8 hours a day, it means:

Time used to the second work = 320 hours / 10
Time used to the second work = 32 hours
Time used to the second work = 32 hours / 8 hours (as each day has 8 hours)
Time used to the second work = 4 days

Now, we know three of the second set can do as much work as four of the first set, taking into account the calculated equivalence, we have:

Work of four workers of first set = Work of three workers of second set
Work of four workers of first set = Equivalence * 4 persons.
Work of four workers of first set = 2 hours /person * 4 persons
Work of four workers of first set = 8 hours.

So, three persons of the second set can make a equivalence of 8 hours. At last, we calculate all the number of workers we need in a regular time:

Number of needed workers in a regular time = (320 hours / 8 hours) * 3 persons.
Number of needed workers in a regular time = 40 * 3 persons
Number of needed workers in a regular time = 120 persons

Remember we need to perform the job not in a regular time, we need to perform it in tenth part of the time, by this reason, we need 10 times the number of people:

Number of needed workers in tenth part of the time = 120 persons * 10
Number of needed workers in tenth part of the time = 1200 persons

With this calculations, you can find the number of men needed to perform a piece of work twice as great in tenth part of the time working 8 hours a day supposing that three of the second set can do as much work as four of the first set is 1200 persons.

4 0

3 years ago