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3 years ago

Simplify please! sqrt(8x^(12) y^(7))

1 answer:

8 0

$\sqrt{8 x^{12} y^{7} } = \sqrt{8} \sqrt{x^{12}} \sqrt{ y^{6}*y } =\sqrt{4} \sqrt{2} \sqrt{(x^{6} )^{2} } \sqrt{ (y^{3})^{2}} \sqrt{y}=2 \sqrt{2} x^{6} y^{3} \sqrt{y} 
$

<span>I hope it helps you :)</span>

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In the Metric System: As you move from a smaller unit to a larger unit, the number of larger units required will be LESS. Theref

3241004551 [841]

Answer:

True

Step-by-step explanation:

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3 years ago

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Compute the probability that exactly 8 of the 20 internet browser users use chrome as their internet browser (to 4 decimals). fo

dybincka [34]

Answer:

2.4%

Step-by-step explanation:

We make use of the binomial probability equation, which is as follows:

P = [n! / (n - r)! r!] p ^ r * q ^ (n - r)

where,

n total number samples = 20

r is the selected number = 8

p, sin este valor no se puede realizar el ejercicio y no lo mencionas, pero encontré una pregunta igual y era de 20.37%, es decir 0.2037

q = 1 - 0.2037 = 0.7963

reemplazando:

P = [20! / ((20 - 8)! * 8!)] * [0.2037 ^ 8 * 0.7963 ^ (20 - 8)]

P = 125970 * 1.92685405*10^-7

P = 0.024 = 2.4%

5 0

3 years ago

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

3 years ago

17. Write the standard form of the equation of the circle

user100 [1]

Answer:

(x -1)^2 + (y -2)^2 = 13

Step-by-step explanation:

The two given points are the end points of a chord, so the center will be on its perpendicular bisector. That is, the center will be at the point of intersection of the given line and the perpendicular bisector of the given chord.

To find that point, we can write the equation of the perpendicular bisector, then solve the simultaneous equations.

The perpendicular bisector can be written as ...

Δx(x -(3+4)/2) +Δy(y -(5+0)/2) = 0

(3 -4)(x -7/2) + (5 -0)(y -5/2) = 0

-x -9 +5y = 0 . . . . . eliminate parentheses

We can add 3 times this equation to the given equation to find the solution for the circle center.

(3x +2y) + 3(-x -9 +5y) = (7) + 3(0)

17y -27 = 7

y = 34/17 = 2

x = 5y -9 = 1

The circle center is (h, k) = (1, 2).

The square of the radius can be found by substituting one of the given points into the formula for the circle. That formula is ...

(x -h)^2 + (y -k)^2 = r^2

Filling in the values for (h, k) and the first given point, we find ...

(4 -1)^2 + (0 -2)^2 = r^2 = 13

The standard form equation for the circle through the given points with the center on the given line is ...

(x -1)^2 + (y -2)^2 = 13

_____

<em>Comments on equation for a line</em>

There are a lot of ways to write the equation of a line. Often, I like to use standard form: ax +by = c. When given two points, (x1, y1) and (x2, y2), this can take the form ...

Δy(x -x1) -Δx(y -y1) = 0

where (Δx, Δy) = (x2 -x1, y2 -y1).

The perpendicular line through some point (h, k) will be of the form ...

Δx(x -h) +Δy(y -k) = 0

Note the change in sign for the second term and the switching of Δx and Δy. This is what makes the slope be the negative reciprocal of the slope of the above line through the two points.

For the perpendicular bisector, the point (h, k) needs to be the midpoint of the segment between (x1, y1) and (x2, y2). That midpoint is the average of the two segment endpoints:

(h, k) = ((x1+x2)/2, (y1+y2)/2)

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3 years ago

Mr. Moore drove the first half of his journey at an average speed of 96 km/h for 2 hours.

Talja [164]

Answer:

bnb

Step-by-step explanation:

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3 years ago