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Wewaii [24]
3 years ago
12

What is the volume of the rectangular prism?

Mathematics
2 answers:
BaLLatris [955]3 years ago
6 0

if you multiply 10.4x5x8 it equals 416 millimeters

Gekata [30.6K]3 years ago
6 0

Step-by-step explanation:

Since the formula of a rectangular prism is when we multiply length times width times height, or 10.4 * 5* 8.

5*8 = 40, and 10.4*40 = 416. (0.4*40 = 16 and 10*40 is 400, so you add them up)

That means that you answer is 416 Millimeters

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With this definition we can conclude that the contrapositive of p→q would be ~q→~p

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If a triangle has sides of length 6 cm and 8 cm, what are the largest and smallest possible lengths for the third side?
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Step-by-step explanation:

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Laurent invested some money in a bank account.
zavuch27 [327]

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Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

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