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3 years ago

Yoo can someone find the missing input pls n thank uuu

Mathematics

1 answer:

swat323 years ago

3 0

Answer:

so x equals 2

Step-by-step explanation:

x =2 because x^3 is the same as 8 and f just means it is a function.

Have a nice day

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Earth is approximately 5x10^9 years old . for which of these ages could this be an a aproxiamation

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In parallelogram MNPQ, side MN is seven inches longer than twice the length of side NP. If the perimeter of MNPQ is 68 inches, t

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Answer: NP = 9 inches

Step-by-step explanation:

In a parallelogram, the opposite sides are equal.

If the length of the longer side is MN, it means that the length of the two opposite longer sides is 2MN.

If the length of the shorter side is NP, it means that the length of the two opposite shorter sides is 2NP.

The perimeter is 2MN + 2NP

If the perimeter of MNPQ is 68 inches, it means that

2MN + 2NP = 68- - - - - - - - - - - -1

In parallelogram MNPQ, side MN is seven inches longer than twice the length of side NP. It means that

MN = 2NP + 7- - - - - - - - - - -2

Substituting equation 2 into equation 1, it becomes

2(2NP + 7) + 2NP = 68

4NP + 14 + 2NP = 68

4NP + 2NP = 68 - 14

6NP = 54

NP = 54/6

NP = 9 inches

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4 years ago

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Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

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