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3 years ago

Can someone answer this question please answer it correctly if it’s correct I will mark you brainliest

Mathematics

1 answer:

vladimir1956 [14]3 years ago

6 0

Answer: 105

Step-by-step explanation: 25 frames is 1 second. The video already played 3 and 2/5 seconds before the player started to count 0, so you can write 3 and 2/5 seconds as an improper fraction. =(5*3)+2 / 5 = 17/5 seconds. Multiply by 25 frame. 17/5 *25 =85 frames. So according to the video counter,after 17/5 seconds,it should count 85 frames.However,at 0 seconds,it indicated a count of 190 frames. then to get the number of frames that were already in count you would subtract 85 frames from the 190 frames.

190-85=105 frames

(Not 100% sure so apologies if I'm wrong)

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

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f(x) = cxⁿ
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Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle H(x) = \sqrt[3]{F(x)}$

Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$