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3 years ago

Help I’m timed I’ll mark u brainliesttt

Mathematics

1 answer:

Zigmanuir [339]3 years ago

4 0

It would be 5sqrt2

a^2 + b^2 = c^2
5^2 + 5^2 = c^2
25 + 25 = c^2
50 = c^2
c = sqrt 50
c = 5sqrt2

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Rudiy27

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3 years ago

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diamong [38]

It would be C as the answers

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2 years ago

Help me fast!!! Best answer gets brain liest

chubhunter [2.5K]

Answer:

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Step-by-step explanation:

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2 years ago

The manager of the BiLo Supermarket in Mt. Pleasant, Rhode Island, gathered the following information on the number of times a c

pishuonlain [190]

Answer: kindly check explanation

Step-by-step explanation:

Given the data :

5 3 3 1 4 4 5 6 4 2 6 6 6 7 1 1 14 1 2 4 4 4 5 6 3 5 3 4 5 6 8 4 7 6 5 9 11 3 12 4 7 6 5 15 1 1 10 8 9 2 12

Frequency and relative frequency table can be found in the attachment below.

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3 up to 6

7 0

3 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.

7 0

3 years ago