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Contact [7]
3 years ago
14

b. From where the shoes spilled (48°N, 161°W) to where they were found on May 22nd 1996 (54°N, 133°W), how many kilometers did t

hey travel? How many days did they take to travel that distance (use April 30 as the date found)? What was their rate of travel in kilometers per hour?
Mathematics
1 answer:
dsp733 years ago
3 0

To find kilometers using latitude and longitude:

distlon = lon2 - lon1   = 133-161 = -23

distlat = lat2 - lat1  = 54-48 = 6

a = (sin(distlat/2))^2 + cos(lat1) * cos(lat2) * (sin(distlon/2))^2  

a = (sin(6/2)^2 + cos(48)*cos(54)* sin(-23/2)^2

c = 2 * arctan2( sqroot(a), sqroot(1-a) )  

d = R * c (where R is the radius of the Earth, 6373 km's)

d = 2052 km's

April 30th to May 22nd is 23 days.

23 days x 24 hours per day = 552 hours.

Rate of travel = 2052 / 552 = 3.717 km per hour ( round answer as needed.)

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Luba_88 [7]
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3 years ago
A solid lies between planes perpendicular to the​ x-axis at x = -2 and x = 2. The​ cross-sections perpendicular to the​ x-axis b
icang [17]

Answer:

V = 128π/3 vu

Step-by-step explanation:

we have that: f(x)₁ = √(4 - x²);  f(x)₂ = -√(4 - x²)

knowing that the volume of a solid is  V=πR²h, where R² (f(x)₁-f(x)₂) and h=dx, then

dV=π(√(4 - x²)+√(4 - x²))²dx;  =π(2√(4 - x²))²dx ⇒

dV= 4π(4-x²)dx , Integrating in both sides

∫dv=4π∫(4-x²)dx , we take ∫(4-x²)dx and we solve

4∫dx-∫x²dx = 4x-(x³/3) evaluated -2≤x≤2 or too 2 (0≤x≤2) , also

∫dv=8π∫(4-x²)dx evaluated 0≤x≤2

V=8π(4x-(x³/3)) = 8π(4.2-(2³/3)) = 8π(8-(8/3)) =(8π/3)(24-8) ⇒

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8 0
3 years ago
What is 363 rounded to the nearest ten and hundred
anyanavicka [17]
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4 0
3 years ago
Read 2 more answers
For the graphed exponential equation, calculate the average rate of change from x = −3 to x = 0.
iragen [17]

Answer:

-\frac{7}{3}

Step-by-step explanation:

To solve this, we are using the average rate of change formula:

m=\frac{f(b)-f(a)}{b-a}

where

m is the average rate of change

a is the first point

b is the second point

f(a) is the function evaluated at the first point

f(b) is the function evaluated at the second point

We want to know the average rate of change of the function f(x)=0.5^x-6 form x = -3 to x = 0, so our first point is -3 and our second point is 0. In other words, a=-3 and b=0.

Replacing values

m=\frac{f(b)-f(a)}{b-a}

m=\frac{0.5^0-6-(0.5^{-3}-6)}{0-(-3)}

m=\frac{1-6-(8-6)}{3}

m=\frac{-5-(2)}{3}

m=\frac{-5-2}{3}

m=\frac{-7}{3}

m=-\frac{7}{3}

We can conclude that the average rate of change of the exponential equation form x = -3 to x = 0 is -\frac{7}{3}

4 0
2 years ago
Which equation is the inverse of y = x2 – 36?
Sholpan [36]
Its B.y=±√​n+36​​​       

i hope that helps 
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3 years ago
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