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2 years ago

What is this for 2nd grade math. 247+?=673

Mathematics

1 answer:

icang [17]2 years ago

7 0

Answer:

426

Step-by-step explanation:

Subtract 247 from 673 to get 426

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The average speed to go by car from home to college is 30 miles per hour. The average speed to go back home from college is 20 m

Levart [38]

Answer:

24 mph for both 60 miles and 40 miles.

Step-by-step explanation:

We need to find how long it takes to get to college and from it.

Speed x time (x) = Distance (60)

30x = 60

x=2

It takes two hours to get to college

20x = 60

x = 3

It takes 3 hours to get back. It's a total of 5 hours, 2 hours at 30mph and 3 at 20. Now we need to find the average

30+30+20+20+20=120

120/5= 24

The average speed for the round trip is 24.

To do the second part, we follow the same process, but replace 60 with 40 for distance.

30x=40

x= 1.33

20x=40

x=2

We can add the total distance and divide by total time to find the average.

The total distance is 80. Time is 3.33

80/3.33=24

The average speed is still 24.

3 0

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

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

4 0

3 years ago

Angles C and E are complementary angles. Angle C measures (4× + 50). Angle E measures 28 degrees. Solve for x.

julsineya [31]

Answer:

x = 3

Step-by-step explanation:

4x + 50 + 28 = 90

4x + 78 = 90

4x = 12

x = 3

7 0