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3 years ago

1/12 divided by 4 =

Mathematics

2 answers:

Ivan3 years ago

7 0

Answer:

1/48 is the correct answer

patriot [66]3 years ago

7 0

1/48 because it’s the same thing as multiplying it by 1/4

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Plz help ASAP with explanation

Aleksandr-060686 [28]

Answer:

The height of the tank in the picture is:

19.5 cm

Step-by-step explanation:

First, to know the height of the tank, we're gonna change the unit of the volume given in liters to cm^3:

1 liter = 1000 cm^3

So:

1.2 liters = 1200 cm^3

Now, we must calculate the height of the tank that we don't know (the other part that isn't with water), to this, we can use the volume formula clearing the height:

Volume of a cube = long * wide * height

Now, we must clear the height because we know the volume (1200 cm^3):

Height = volume of a cube / (long * wide)

And we replace:

Height = 1200 cm^3 / (12 cm * 8 cm)
Height = 1200 cm^3 / (96 cm^2)
Height = 12.5 cm

Remember this is the height of the empty zone, by this reason, to obtain the height of the whole tank, we must add the height of the zone with water (7 cm) that the exercise give us:

Heigth of the tank = Height empty zone + height zone with water
Heigth of the tank = 12.5 cm + 7 cm
Heigth of the tank = 19.5 cm

In this form, we calculate the height of the tank in 19.5 cm.

6 0

3 years ago

Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.

kupik [55]

The distance between some point $(x,y,z)$ and the origin is given by

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

so this is the function we're trying to minimize. But notice that $f(x,y,z)$ and $f(x,y,z)^2$ attain their critical points at the same $(x,y,z)$ , so we can solve the same problem by minimizing $x^2+y^2+z^2$ instead.

So let's take the Lagrangian to be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)$

with partial derivatives (set equal to 0)

$L_x=2x+a\lambda=0$
$L_y=2y+b\lambda=0$
$L_z=2z+c\lambda=0$
$L_\lambda=ax+by+cz-d=0$

Now, notice that

$aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0$
$\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}$

and we can use this to solve for $x,y,z$ . We get

$x=\dfrac{ad}{a^2+b^2+c^2}$
$y=\dfrac{bd}{a^2+b^2+c^2}$
$z=\dfrac{cd}{a^2+b^2+c^2}$

At this critical point, we get a minimum distance of

$\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}$

8 0

3 years ago

Need help. If u answer fast or if u are first i will give u brainliest

aksik [14]

Answer:

It is correct.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Set-up an equation to model the problem, then solve that equation:

ivolga24 [154]