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fgiga [73]
3 years ago
9

M is directly proportinal to r^2 when r = 2 m= 14 work out the value of m

Mathematics
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

14?

Step-by-step explanation:

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Harvey rides his bike at an average speed of 12 miles per hour. In other words, he rides 12 miles in 1 hour, 24 miles in 2 hours
timama [110]

Answer:

Step-by-step explanation:

Harvey rides his bike at an average speed of 12 miles in 1 hour, 24 miles in 2 hours, and so on. Let h be the number of hours he rides and d be distance traveled. Write the equation for the relationship between distance and time in point-slope form.

----

You have three points relating time and distance:

(1,12),(2,24),(3,36)

-----

The slope for your equation would be (24-12)/(2-1) = 12/1 = 12

-------------------------------

The intercept is the distance you would travel in no time; that would be zero.

---------------------------------

The equation is: distance = 12(time)

d = 12t

4 0
3 years ago
I need the answer to number 4
zloy xaker [14]

Answer: 6.3

Step-by-step explanation

I did this today and it's easy but you have to show a lot of work well I did but you don't have to you can just circle it in

5 0
3 years ago
If the diameter of the circle below is 16 in., what is the area of the circle?
mote1985 [20]

Answer:

64π inches² (approximately 201.06 inches²)

Step-by-step explanation:

A=\pi r^2 where r is the radius

To find the radius, divide the diameter by 2

16 ÷ 2 = 8 inches

Plug in 8 as the radius

A=\pi (8)^2\\A=64\pi

A=201.06 (approximately)

Therefore, the area of the circle is 64π inches², or approximately 201.06 inches².

I hope this helps!

8 0
3 years ago
Read 2 more answers
Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

Answer:

The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

  • x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
4 years ago
A dilation has center (0, 0). Find the image of each point for the given scale factor.
balu736 [363]

Here we have a point and dilation factor and the dilation factor or the scale factor is 12 . SO we need to multiply each coordinate by 12, that is 0 times 12=0 and 4 times 12=48 .

So the image of point after dilation is (0,48) .

7 0
3 years ago
Read 2 more answers
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