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mr_godi [17]
3 years ago
14

Please only say something if you really know it

Mathematics
1 answer:
Agata [3.3K]3 years ago
7 0

Answer:

x axis

Step-by-step explanation:

it will never cross the x axis

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10.
harina [27]
The answer is D.. :)
7 0
2 years ago
Gill Rolls a number Cube 78 times how many times can he expect to roll a number greater than one
saul85 [17]
Argument
The only number he can't roll is the number one. There are 5 numbers that he can roll.

The probability of success therefore is 5/6.

Formula
Number of successful rolls = probability of success * total number of rolls.
Total number of rolls = 78
Probability = 5/6

Number of successful Rolls = 5/6 * 78 
Number of successful Rolls = 390/6
Number of successful Rolls = 65

Answer
He should roll a number greater than one 65 times <<<<< answer

5 0
3 years ago
Rank the following values from largest to smallest.
Scorpion4ik [409]

Answer:

Population Of the United States

Weight of a small car

Length of a bed

Weight of an average man

7 0
3 years ago
PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. <br> IM GIVING 40 POINTS !! DONT SKIP :((.
oksian1 [2.3K]

m = 1.49

b = 5.50

equation y = 1.49x + 5.50

Non-proportional

Positive Slope

8 0
3 years ago
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
3 years ago
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