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2 years ago

8

M∠LONm, angle, L, O, N is a straight angle.

1 answer:

Ne4ueva [31]2 years ago

5 0

Answer: 41

Step-by-step explanation:

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Find the gradient & y-intercept for each equation of a line.

Hitman42 [59]

Answer:

4.56×223458 ×4568÷67765

6 0

2 years ago

there are 120 students in a school marching band. they march in an array with the same number of students in each row. what are

Sloan [31]

Answer:

20 X 6

Step-by-step explanation:

20 X 6 = 120 I'm not smart but this is pretty easy

5 0

2 years ago

I need help with an Algebra question. The question is in the picture. Thank you <3

olchik [2.2K]

The domain is the set of allowed inputs, in this case t values. The smallest t value allowed is t = 0. The largest is t = 165. So that's why the domain is $0 \le t \le 165$

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The range is $0 \le H \le 40000$ since H = 0 is the smallest output of the function and H = 40,000 is the largest output. Like the domain, the range is the set of possible outputs of a function.

4 0

2 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

a point on the rim of a wheel moves with a velocity of 80 feet per second. What is the angular velocity of the point if the diam

valentina_108 [34]

The angular velocity is 16 radians per second

<u>Solution:</u>

The angular velocity of point is given by formula:

$w = \frac{v}{r}$

Where,

w is the angular velocity in radians

v is the linear velocity

r is the radius

Given that diameter of wheel is 10 feet

$radius=\frac{diameter}{2}$

$radius = \frac{10}{2}=5$

Thus radius is 5 feet

A point on the rim of a wheel moves with a velocity of 80 feet per second

Linear velocity = v = 80 feet per second

Therefore, angular velocity is given as:

$w = \frac{80}{5} = 16$

Thus angular velocity is 16 radians per second

6 0

2 years ago