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3 years ago

What is the approximate number of hours in 72 days on Saturn

Mathematics

2 answers:

amid [387]3 years ago

4 0

The answer would be about 743 hours in 72 days on saturn

Benjamin Courtney

2 years ago

Thx bro.

Anon25 [30]3 years ago

3 0

A planet's day is the time it takes the planet to rotate or spin once on its axis. Saturn rotates faster than Earth so a day on Saturn is shorter than a day on Earth. A day on Saturn is 10.656<span> hours long while a day on Earth is </span>23.934<span> hours long.</span><span />

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Melina graphed a line which passes through the points (-4, 2) and (6, 8). What is the slope of the line?

Ulleksa [173]

Answer:

The answer is

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4, 2) and (6, 8)

The slope is

$m = \frac{8 - 2}{6 + 4} = \frac{6}{10} = \frac{3}{5} \\$

We have the final answer as

$\frac{3}{5} \\$

Hope this helps you

7 0

3 years ago

Suppose on a certain test, fifty students earn a perfect score and fifty students earn a zero. determine the mean grade and stan

kodGreya [7K]

Answer:

Mean=50.

standard deviation=50.

Step-by-step explanation:

Let the perfect score be 100.

The mean(average) of a data is given by: $Mean=\frac{sum of data points}{number of data points}$

Here the number of data points are 100. out of which 50 attains a value 100,and 50 attains value 0.

so, sum of data points=50×100+50×0=5000.

$Mean=\frac{5000}{100}$

Mean=50.

"Now the standard deviation of data points are calculated by firstly subtracting mean from every entry and then square the number and take it as new entry and calculate the mean of the new data entry and lastly taking the square root of this new mean".

Here if 50 is subtracted from each entry the new entry will have 50 entries as '50' and 50 entries as '-50'.

next on squaring we will have all the 100 entries as '2500'.

now the mean of these entries is: $\frac{2500\times100}{100}$

=2500

taking it's squareroot we have $\sqrt{2500}=50$

Hence, standard deviation=50.

5 0

3 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.