Problem 1

The formula to use is
s = r*theta

where,
s = arc length
r = radius
theta = angle in radians
Note: if the angle is not in radians, you have to convert it over to radians. However, in this case we are told the angle is "5pi/4 radians". So no conversion is needed.

In this case,
s = unknown (this is what we want to solve for)
r = 34
theta = 5pi/4
keep in mind that 5pi/4 is the same as (5/4)*pi

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Plug the value of r and theta into the formula to get...
s = r*theta
s = 34*(5pi/4)
s = (34*5/4)*pi
s = 42.5*pi
s = 42.5*3.14
s = 133.45

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Final Answer: 133.45

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Problem 2

sin(angle) = opposite/hypotenuse
sin(61.7) = 8/y
y*sin(61.7) = 8
y = 8/sin(61.7)
y = 9.08598042654456 ... use a calculator
y = 9 ... rounding to the nearest whole number

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Final Answer: 9

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Problem 3

We need to know the length of BC. Let's call this x for now. Use the pythagorean theorem to find x

x^2 + 14^2 = 50^2
x^2 + 196 = 2500
x^2 + 196-196 = 2500-196
x^2 = 2304
sqrt(x^2) = sqrt(2304)
x = 48

So BC = 48 units long
Now we can compute the tangent of angle B

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tan(angle) = opposite/adjacent
tan(B) = AC/BC
tan(B) = 14/48
tan(B) = 7/24

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Final Answer: 7/24

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Problem 4

In this current stage, I cannot answer this question because the image isn't showing up. On my end, all I see is a black rectangle with no triangle showing. It seems like some kind of glitch is happening. Please repost this image. Thank you.

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Problem 5

There are multiple answers here so it seems like there should be a restriction. Does it state what the restriction must be? Please let me know. Thank you.

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Problem 6

Rule:
sin(x) = cos(90-x)
where x is an angle in degrees

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Using that rule mentioned above, we can replace the "sin(x)" term with "cos(90-x)" then solve for x

cos(63) = sin(x)
cos(63) = cos(90-x)
both arguments must be equal, so 63 must be equal to 90-x

63 = 90-x
63+x = 90-x+x
x+63 = 90
x+63-63 = 90-63
x = 27

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Final Answer: 27

5 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height

Ilia_Sergeevich [38]

Answer:

$\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s$

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

$\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}$

From this point on, you have two possibilities, either you find each individual $y_{f}, y_{0}, t_{f}, t_{0}$ and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.