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Molodets [167]
3 years ago
12

Please help trig functions !!!! ASAP

Mathematics
2 answers:
mojhsa [17]3 years ago
5 0
Csc means cosecant and it is the ratio of the hypotenuse / opposite.
With angle B, the hypotenuse = 10 and the opposite side = 6.
So, the cosecant of Angle B = 10 / 6 = 1.6666666...
To see all the trigonometric ratios, click here:
http://www.1728.org/trigcalc.htm


goblinko [34]3 years ago
3 0
We will use the ratio of tanget which is the opposite over the adjecent
The opposite side of angle B is 6
The adjecent side of angle B is 8
Since you are trying to find angles we will use the universe trig functions

\tan {}^{ - 1} ( \frac{6}{8} )  = 36.9
The measurementof angle B is 36.9 degrees.
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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

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Answer: skrt skrt



Step-by-step explanation:


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Answer:

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Step-by-step explanation:

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This is found by multiplying 500 (starting number of candy) and .64 (percentage divided by a hundred). Thjs would guve you 320, which you would then subtract from the starting number of candy (500) to get 180. 180 is Yolanda's number of candy, which gives you the answer.
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