If AB represents 50% , what is the length of a line segment that is 100%

Determine the values of the variables in isosceles trapezoid CHLE below.

Alex73 [517]

Answer:

The values of variables x and m are 11 and 17

Step-by-step explanation:

The question has missing details as the diagram of the trapezoid isn't attached.

(See attachment).

Given that trapezoid CHLE is isosceles then the angles at the base area equal (4x)

And

The angles at the top are also equal

8m = 11x + 15

At this point, the four angles in the trapezoid are 8m, 11x + 15, 4x and 4x..

The sum of interior= 360

So,

11x + 15 + 8m + 4x + 4x = 360

Collect like terms

11x + 4x + 4x + 8m = 360 - 15

19x + 8m = 345

Substitute 11x + 15 for 8m

19x + 11x + 15 = 345

30x + 15 = 345

30x = 345 - 15

30x = 330

Divide through by 30

30x/30 = 330/30

x = 11

Recall that 8m = 11x + 15;

8m = 11(11) + 15

8m = 121 + 15

8m = 136

Divide through by 8

8m/8 = 136/8

m = 17

Hence, the values of variables x and m are 11 and 17

7 0

3 years ago

The difference of 7 and a number x is 10

Oksi-84 [34.3K]

The answer to the question

5 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

Sines/cosines please help!!!!

AleksandrR [38]

I can help explain it if you PM me.

8 0

3 years ago

2. Janine walks on a circular track whose radius is 60 meters. If she walks a total of 100 meters, which of the

katrin2010 [14]

Answer:

0.53πrad

Explanations:

Given the radius of the circular track = 60metres

If she walks a total of 100 meters, the length of the arc of the circle = 100metres

To calculate the radian angle she rotates about the center of the track, we will use the formula for calculating the length of an arc

L = θ/360° × 2πr

100 = θ/360× 2π(60)

36000 = 120π × θ

36000 = 376.8θ

θ = 36000/376.8

θ = 95.5°

Since 180° = πrad

95.5° = x

x = 95.5π/180

x = 0.53π rad

6 0

3 years ago