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2 years ago

Find NP from the Rectangle

Mathematics

1 answer:

antoniya [11.8K]2 years ago

4 0

Answer:

9.16

Step-by-step explanation:

by using pythagorean theorem,

given that PQ=4 and NQ = 2×5 = 10

NP²= NQ²-PQ²

= 10²-4²

= 100-16

= 84

NP = √84 = 9.16

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Answer:

1 pair of boots and 6 sweaters

Step-by-step explanation:

70+20s = 135

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20s = 135

divide 20 both sides

s = 6.75

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2 years ago

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3 years ago

Use the theoretical method to determine the probability of the following event. A randomly selected person has a birthday in Nov

MakcuM [25]

Answer:

There is an 8.22% probability that a randomly selected person has a birthday in November.

Step-by-step explanation:

The theoretical method to find the probability is the division of the number of desired outcomes by the number of total outcomes.

A randomly selected person has a birthday in November

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There are 30 days in november, so the number of desired outcomes is 30.

So the probability is

$P = \frac{30}{365} = 0.0822$

There is an 8.22% probability that a randomly selected person has a birthday in November.

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3 years ago

What is an equivalent ratio to 8:20 and<br> 2:5?

coldgirl [10]

Answer:

4:10.

Step-by-step explanation:

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3 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much

Y_Kistochka [10]

Answer:

x = ∛ 2*V/5

y = ∛ 2*V/5

h = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h where h = V / x*y

As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h

According to this:

Ct (cost of aquarium ) = cost of the base + cost of the sides

cₐ ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h

C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or

C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(y) = 5*c₁*x - 2*c₁*V/y²

C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²

or 5*y - 2*V/x² = 5*x - 2*V/y²

(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²

5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)

by symmetry x = y

Then using x = y and plugging that value on the derivatives

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(x) = 5*c₁*x - 2*c₁*V/x²

C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0

5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5

x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25

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3 years ago