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3 years ago

How do I know when to use Sin, Cos or Tan on Trig Ratios?

Mathematics

1 answer:

Sergio039 [100]3 years ago

4 0

Answer:

Usually in the equation they use Sin, Cos, or tan. Although, If the point in the graph starts at 0 is is sin and anything else would be cos. Tan is the sort of vertical squiggly line.

Step-by-step explanation:

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Liam descended from the summit of K2 (elevation= 28,251.31 feet) to an elevation of 23,201.06 feet. How many feet did Liam desce

saul85 [17]

Laim descended 5,050.25 feet and his elevation changed by -5,050.25.

6 0

4 years ago

Jeanette purchased a concert ticket on a web site. The original price of the ticket was $75. She used a coupon code to receive a

DedPeter [7]

Answer:

<u>Jeanette’s ticket was less than the original price by 12 percent</u>

Step-by-step explanation:

1. Let's calculate the 20% discount price Jeanette received:

75 - 0.2 * 75 = 75 - 15 = $ 60

The discounted price is $ 60

2. Let's calculate the 10% service fee to the discounted price:

60 + 0.1 * 60 = 60 + 6 = $ 66

The price including the service fee is $ 66

3. Finally, let's calculate the less percentage than the original price:

Let's use the Direct Rule of Three, this way:

Price Percentage

75 100

66 x

**************************

75x = 66 * 100

75x = 6,600

x = 6,600/75

x = 88

Less percentage = 100 - 88 = 12%

<u>Jeanette’s ticket was less than the original price by 12 percent</u>

3 0

3 years ago

Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.