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3 years ago

What is the square route of 5?

Mathematics

1 answer:

denis-greek [22]3 years ago

6 0

Square root are the same 2 numbers multiplied.

Ex: 8*8=64 (Square root is 8)

9*9=81 (Square root is 9)

2.2360679775*2.2360679775=5

Answer=2.2360679775

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Find the median and mean of the data which reflects the best measure of the center

galben [10]

Answer:

The mean reflects the best measure of the center

Step-by-step explanation:

12, 13, 13, 14, 21, 22, 23

Median: 14

Mean: 16.86

The mean reflects the best measure of the center

7 0

3 years ago

GEOMETRY

Katarina [22]

SA = 2(LW + WH + LH)
SA = 2[8(6) + 6(5) + 8(5)]
SA = 2(48 + 30 + 40)
SA = 2(118)
SA = 236 square yards

5 0

2 years ago

Steve cycled 35 miles in 2 1/2 hours. What is the average speed that Steve traveled? miles per hour

Evgesh-ka [11]

Answer:

14 miles per hour

Step-by-step explanation:

The average speed would be the total distance divided by the total time = 35 miles divided by 2.5 hours is 14 miles per hour

5 0

3 years ago

Sea level = 0

FromTheMoon [43]

Answer:

c. -70

Step-by-step explanation:

subtract 50 from 120 you get 70, and since we are below sea level it is negative

7 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago