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3 years ago

9

Because of a problem in the program, the timer in a video player did not begin counting until the video had been playing for sev

eral seconds. The player began counting at 0 seconds even though the video had already played 190 frames The video plays 252525 frames per second. How many frames had the video already played when the time was equal to -3 3/5

1 answer:

Bess [88]3 years ago

3 0

Answer:

105 frames

Step-by-step explanation:

Given that 25 frames are played per second

25 frames= 1 sec

The video already played 3 and 2/5 seconds before the player started to count 0

Write 3 and 2/5 seconds as an improper fraction

=(5*3)+2 / 5 = 17/5 seconds

Multiply by 25 frames

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.Thus,to get the number of frames that were already in count you subtract 85 frames from the 190 frames.

190-85=105 frames.

Step-by-step explanation:

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The width of a box in the shape of a rectangular prism is 4 inches more than the height h . The length is the difference of 9 in

Doss [256]

Answer:

$(a)$ $V = -H^3+ 5H^2 + 36H$

$(b)$ $Length = 5in; Width = 9in; Height = 5in$ or $Length = 3in; Width = 10in; Height = 6in$

Step-by-step explanation:

Let:

$W=Width; L = Length; H = Height$

Given

$W = 4 + H$

$L = 9 - H$

Solving (a): The polynomial for the volume in terms of height

Volume (V) is calculated using:

$V = L * W * H$

$V = (9 - H) * (4 + H) * H$

Expand

$V = (9 - H) * (4H + H^2)$

Open brackets

$V = 36H + 9H^2 - 4H^2 - H^3$

$V = 36H + 5H^2 - H^3$

Rewrite as:

$V = -H^3+ 5H^2 + 36H$

Solving (b): The dimensions when Volume = 180

Substitute 180 for V in: $V = -H^3+ 5H^2 + 36H$

$180 = -H^3+ 5H^2 + 36H$

Rewrite as:

$H^3 - 5H^2 - 36H + 180 = 0$

Factorize:

$H^2(H - 5) - 36(H - 5) = 0$

$(H^2 - 36)(H - 5) = 0$

Express $H^2 - 36$ as difference of two squares

$(H - 6)(H + 6)(H - 5) =0$

This gives:

$H - 6 =0==> H = 6$

$H + 6 =0==> H = -6$

$H - 5 =0==> H = 5$

Since the height can not be negative, then H = 5 or H = 6

Recall that:

$W = 4 + H$

$L = 9 - H$

When H = 5

$W = 4 + 5 = 9$

$L = 9 - 5 = 4$

When H = 6

$W = 4 + 6 = 10$

$L = 9 - 6 = 3$

Hence, the dimensions of the box are:

$Length = 5in; Width = 9in; Height = 5in$

or

$Length = 3in; Width = 10in; Height = 6in$

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denpristay [2]

Answer:

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Step-by-step explanation:

1/4 × 7/9 = 7/36 Hope that helps!

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