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Novosadov [1.4K]
2 years ago
8

Help me answer this question

Mathematics
2 answers:
vovangra [49]2 years ago
8 0

Answer: I think 128 cubes

Step-by-step explanation:

We are given: Volume of rectangular prism = 2 units³ and Cube with edge = 1/4 unitWe must find: The number of cubes that can fill the prismFirstly, find the volume of 1 cube:Volume = 1/4 x 1/4 x 1/4Volume = 1/64 units³Next find the number of cubes that can go into the prism:Number of cubes = 2 ÷ 1/64Number of cubes = 128Therefore 128 cubes can fit in.

ad-work [718]2 years ago
4 0

Answer:

128 cubes can fit in.

Step-by-step explanation:

We are given: Volume of rectangular prism = 2 units³ and Cube with edge = 1/4 unit

We must find: The number of cubes that can fill the prism

Firstly, find the volume of 1 cube:

Volume = 1/4 x 1/4 x 1/4

Volume = 1/64 units³

Next find the number of cubes that can go into the prism:

Number of cubes = 2 ÷ 1/64

Number of cubes = 128

Therefore 128 cubes can fit in.

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Show that the function f(x)=sin3x + cos5x is periodic and it’s period.
lbvjy [14]

The period of f(x) is \boxed{2\pi}.

Recall that \sin(x) and \cos(x) both have periods of 2\pi. This means

\sin(x + 2\pi) = \sin(x)

\cos(x + 2\pi) = \cos(x)

Replacing x with 3x, we have

\sin(3x + 2\pi) = \sin\left(3 \left(x + \dfrac{2\pi}3\right)\right) = \sin(3x)

In other words, if we change x by some multiple of \frac{2\pi}3, we end up with the same output. So \sin(3x) has period \frac{2\pi}3.

Similarly, \cos(5x) has a period of \frac{2\pi}5,

\cos(5x + 2\pi) = \cos\left(5 \left(x + \dfrac{2\pi}5\right)\right) = \cos(5x)

We want to find the period p of f(x), such that

f(x + p) = f(x)

\implies \sin(3x + p) + \cos(5x + p) = \sin(3x) + \cos(5x)

On the left side, we have

\sin(3x + p) = \sin(3x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \sin(3x+2\pi) \cos(p-2\pi) + \cos(3x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \sin(3x) \cos(p-2\pi) + \cos(3x) \sin(p - 2\pi)

and

\cos(5x + p) = \cos(5x + 2\pi + p - 2\pi) \\\\ ~~~~~~~~ = \cos(5x+2\pi) \cos(p-2\pi) - \sin(5x+2\pi) \sin(p-2\pi) \\\\ ~~~~~~~~ = \cos(5x) \cos(p-2\pi) - \sin(5x) \sin(p-2\pi)

So, in terms of its period, we have

f(x) = \sin(3x) \cos(p - 2\pi) + \cos(3x) \sin(p - 2\pi)  \\\\ ~~~~~~~~ ~~~~+ \cos(5x) \cos(p - 2\pi) - \sin(5x) \sin(p - 2\pi)

and we need to find the smallest positive p such that

\begin{cases} \cos(p - 2\pi) = 1 \\ \sin(p - 2\pi) = 0 \end{cases}

which points to p=2\pi, since

\cos(2\pi-2\pi) = \cos(0) = 1

\sin(2\pi - 2\pi) = \sin(0) = 0

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vladimir2022 [97]

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Step-by-step explanation:

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3 years ago
Stephanie took a science exam that had two parts. Part 1: She got 16 questions correct, which was 45 of the number of part 1 que
nasty-shy [4]

Answer:

44

Step-by-step explanation:

Let x and y be the total number of questions in part 1 and part 2 exams respectively.

For part 1:

Number of correct question = 4/5 of the total number of question in part 1

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For part 2:

Number of corrected question = 2/3 of total number of question in part 2

\Rightarrow 16=\frac{2}{3}\times x

\Rightarrow x=24

So, the total umbers of the questions on Stephanie's exam

=x+y

=20+24

=44

Hence, the total umbers of the questions on Stephanie's exam was 44.

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How do you write 3x-y=14 in slope intercept form
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