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Dennis_Churaev [7]

1 year ago

8

The drama club meets in the school auditorium every 8 days, and the choir meets there every 6 days. If the groups are both meeti

ng in the auditorium today, then how many days from now will they next have to share the auditorium?

1 answer:

kondaur [170]1 year ago

4 0

You need to find the least common multiple of 8 and 6.

$8=2^3\\6=2\cdot3\\\\\text{lcm}(8,6)=2^3\cdot3=24$

Therefore, the answer is 24 days.

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pshichka [43]

Answer:

point symmetry if no symmetry isn't an option as it looks grayed out and not clickable

5 0

1 year ago

2. Al and Ike decide to drive to Sandy Shores Lake. As they leave home in Summerville, they notice that the car odometer reads 1

Lilit [14]

When they leave home, odometer reads 12,544.....when they arrive in Winterton, the odometer reads 12,671.....this means they have already driven (12,671 - 12,544) = 127 miles.....and now they have 48 more miles to go......so the trip to Sandy Shores from Summerville is (127 + 48) = 175 miles

6 0

2 years ago

Use lagrange multipliers to find the point on the plane x â 2y + 3z = 6 that is closest to the point (0, 2, 4).

Arisa [49]

The distance between a point $(x,y,z)$ on the given plane and the point (0, 2, 4) is

$\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}$

but since $\sqrt{f(x,y,z)}$ and $f(x,y,z)$ share critical points, we can instead consider the problem of optimizing $f(x,y,z)$ subject to $x-2y+3z=6$ .

The Lagrangian is

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6$

Solve for $\lambda$ :

$x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6$
$\implies2=7\lambda\implies\lambda=\dfrac27$

which gives the critical point

$x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7$

We can confirm that this is a minimum by checking the Hessian matrix of $f(x,y,z)$ :

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.

At this point, we get a distance from (0, 2, 4) of

$\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}$

8 0

2 years ago

a flag pole that is 8 feet tall casts a shadow that is 6 ft long a nearby tree casts a shadow that is 21 ft long. How tall is th

Andru [333]

Answer:

The tree is 28 feet tall.

Step-by-step explanation:

It is given that:

Height of flag pole = 8 feet

Shadow of flag pole = 6 feet

Ratio of height to shadow = 8 : 6

Shadow of nearby tree = 21 feet

Height of nearby tree = x

Ratio of height to shadow = x : 21

8 : 6 :: x : 21

Product of means = Product of extremes

6x = 8*21

6x = 168

Dividing both sides by 6

$\frac{6x}{6}=\frac{168}{6}\\x=28$

Therefore,

The tree is 28 feet tall.

7 0

2 years ago

Hello, please help!! Thank you

Art [367]

I gave steps in the picture how to solve it was much easier to write it out then type it. hope this helps!

3 0

2 years ago