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3 years ago

If y varies directly as x, and y is 20 when x is 4, what is the constant of variation for this relation?

Mathematics

1 answer:

yuradex [85]3 years ago

3 0

Y
y=ax
where a is a constant of proportionality;
y=20, x=4
20=4a
a=20/4
a=5
y=5x
The constant of variation is 5

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Which expression is equivalent to y · 48?

gtnhenbr [62]

Answer:

None of the above

Step-by-step explanation:

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2 years ago

PLEASEEEEEEEEEEEEEEEEEE URGENT

jeyben [28]

Option C:

f(x) = 5x + 7 is the function of the input-output table.

Solution:

Option A: f(x) = 6x + 9

Input x = 1 in the above equation.

f(1) = 6(1) + 9

= 6 + 9

f(1) = 15

But the output is 12 in the table.

So, it is not the function of the table.

Option B: f(x) = 7x + 5

Input x = 1 in the above equation.

f(1) = 7(1) + 5

f(1) = 12

Input x = 5 in the above equation.

f(5) = 7(5) + 5

f(5) = 40

But the output is 32 in the table.

So, it is not the function of the table.

Option C: f(x) = 5x + 7

Input x = 1 in the above equation.

f(1) = 5(1) + 7

f(1) = 12

Input x = 5 in the above equation.

f(5) = 5(5) + 7

f(5) = 32

Input x = 10 in the above equation.

f(10) = 5(10) + 7

f(10) = 57

Input x = 5 in the above equation.

f(15) = 5(15) + 7

f(15) = 82

All outputs are correct for the given input.

Hence it is the function of the table.

Option D: f(x) = 12x + 1

Input x = 1 in the above equation.

f(1) = 12(1) + 1

f(1) = 12

Input x = 5 in the above equation.

f(5) = 12(5) + 1

f(5) = 61

But the output is 32 in the table.

So, it is not the function of the table.

Hence Option C is the correct answer.

f(x) = 5x + 7 is the function of the input-output table.

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3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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Mkey [24]

Answer:

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Step-by-step explanation:

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3 years ago

Where do you place the parentheses in the expression 4+3^2•5-2 to equal 31

Pani-rosa [81]

4+3^2•5-2

$4+3^2\cdot(5-2)=4+9\cdot3=4+27=31$

Used PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

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3 years ago