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stealth61 [152]
3 years ago
9

In ΔEFG , what is the approximate length of EF if the measure of angle G is 50° and FG is 10 cm?

Mathematics
1 answer:
swat323 years ago
6 0
Hey again, you can use trigonometric ratios for this one. You are trying to find the side length opposite of angle G, when you have the adjacent side length value known. You can use TOA, or tan((opposite/adjacent)) . Tan(50) = x/10 

X=  11.92 cm
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Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
Aaron is using a special beaker in science that is a cylinder with a radius of 2 cm and a height of 3 cm on top of a cylinder th
Vilka [71]
Since in the above case, the beaker has two sections each with different radius and height, we will divide this problem into two parts.

We will calculate the volume of both the beakers separately and then add them up together to get the volume of the beaker.

Given, π = 3.14

Beaker 1:

Radius (r₁) = 2 cm
Height (h₁) = 3 cm

Volume (V₁) = π r₁² h₁ = 3.14 x 2² x 3 = 37.68 cm³

Beaker 2:

Radius (r₂) = 6 cm
Height (h₂) = 4 cm
Volume (V₂) = π r₂² h₂ = 3.14 x 6² x 4 = 452.16 cm³

Volume of beaker = V₁ + V₂ = 37.68 + 452.16 = 489.84 cm³


7 0
3 years ago
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3. Write an expression that would help you
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Answer:

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1 year ago
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Answer:

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height (h) = 36 units

Plug in the values and find the volume in terms of π

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