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3 years ago

Hola alguien me puede decir cual es la respuesta

Mathematics

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

D, Option A $\sqrt{3}$

Step-by-step explanation:

CE = $\sqrt{3}$ (by using pythagorean theorem

A = 1/2 bh

A = 1/2 (2)( $\sqrt{3}$ ) = $\sqrt{3}$

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a circle has an area of 225 square inches. if the radius of the circle is multiples by 6, then what is the new area of the circl

Lesechka [4]

Answer:

D: 8,100π in²

Step-by-step explanation:

The formula for area of a circle is

A = πr²

The radius is squared, so if the radius or a new circle is 6 times larger (multiplied by 6), the difference in area will be 36 times greater. So the area is

225π(36) = 8,100π

3 0

2 years ago

I WILL GIVE BRAINLIEST

marshall27 [118]

Answer:

Step-by-step explanation:

60 is ur answer or 36 is ur answer

Pls do Mark it

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3 years ago

A guide wire of length 108 meters runs from the top of an antenna to the ground. If the angle of elevation to the top of the ant

netineya [11]

Answer: Approximately 72.69 meters

Step-by-step explanation:

Antenna height = h

$sin(42.3)=\frac{opposite}{hypotenuse} =\frac{h}{108} \\\\108*sin(42.3)=h\\\\h=72.685$

5 0

2 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$