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3 years ago

Damien is buying tickets to a movie. The movie he and his friends want to watch cost $4.85 per ticket. Damien has $32.00 total.

Mathematics

2 answers:

skad [1K]3 years ago

7 0

Answer:

Step-by-step explanation:

Setup a equation.

32=4.85x (x equals amount of people)

Then divide both sides.

x=6.60

Obviously you can't split a person into .6, so the greatest amount of tickets is 6

salantis [7]3 years ago

4 0

X=greatest number of tickets Damien can afford to buy

x=32/4.85
x=6.56
x~6
Damien can buy 6 tickets
6•4.85=31.96$

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Step-by-step explanation:

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3 years ago

What is the explicit rule for this sequence? -5, 15, – 45, 135, ...

hoa [83]

Answer:

Step-by-step explanation:

In a geometric sequence, consecutive terms differ by a common ratio. The formula for determining the nth term of a geometric progression is expressed as

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4 years ago

What is the largest possible integral value in the domain of the real-valued function

kotegsom [21]

Answer:

Max Value: x = 400

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

Antiderivatives
Integral Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$
Integration Method: U-Substitution
[Integration] Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C$

Step-by-step explanation:

Step 1: Define

$f(x) = \frac{1}{\sqrt{800-2x} }$

Step 2: Identify Variables

Using U-Substitution, we set variables in order to integrate.

$u = 800-2x\\du = -2dx$

Step 3: Integrate

Define: $\int {f(x)} \, dx$
Substitute: $\int {\frac{1}{\sqrt{800-2x} } } \, dx$
[Integral] Int Property: $-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx$
[Integral] U-Sub: $-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du$
[Integral] Rewrite: $-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du$
[Integral - Evaluate] Reverse Power Rule: $-\frac{1}{2}(2\sqrt{u}) + C$
Simplify: $-\sqrt{u} + C$
Back-Substitute: $-\sqrt{800-2x} + C$
Factor: $-\sqrt{-2(x - 400)} + C$

Step 4: Identify Domain

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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