To solve for the surface area of the pyramid, we make use
of the formula:
A= l w + l [sqrt ((w / 2)^2 + h^2)] + w [sqrt ((l / 2)^2 + h^2))
where,
l and w are the base of the pyramid = 100 mm
h is the height of the pyramid = 75 mm
Substituting the given values into the equation:
A= 100 * 100 + 100 [sqrt ((100 / 2)^2 + 75^2)] + 100 [sqrt ((100
/ 2)^2 + 75^2))
A = 10,000 + 100 (sqrt 2575) + 100 (sqrt 2575)
A = 20,148.90 mm^2
Therefore the surface area of the pyramid is about 20,149
mm^2.
Answer is 19/60. The gcf of both numbers is 5. So divide both top and bottom by 5
95 divided by 5 is 19
300 divided by 5 is 60
19/60 can't be divided any further so that is the answer
Answer:
0.33
Step-by-step explanation:
Given the following :
P(speeding) = p(s) = 0.75
P(being stopped) = p(t)
P(speeding and gets stopped) = p(s n t) = 0.25
Find the probability that he is stopped, given that he is speeding is written as P(t | s) ;
P(t | s) = p(s n t) / p(s)
P(s n t) = 0.25
P(s) = 0.75
Hence,
P(t | s) = 0.25 / 0.75
P(t | s) = 0.33
Answer:
40,608,000 km or 4.0608×
km
Step-by-step explanation:
Speed(km/h) = Distance(km)/Time(Hours)
Distance = Speed × Time
6 days = 24 × 6 = 144 hours
282,000 km/h × 144 hours = 40,608,000 km
I think of these in terms of "short side", "long side" and "hypotenuse." The proportions will be true that equate the ratios of corresponding sides.
I is (long side of CDB)/(short side of CDB) = (long side of BDA)/(short side of BDA)
These are corresponding sides, so the proportion is <em>true</em>.
II is (hypotenuse of CBA)/(long side of CBA) = (long side of CBA)/(hypotenuse of CBA)
This proportion equates a value to its inverse, so is <em>false</em> (since the triangle has non-zero area)
III is (long side of CBA)/(short side of CBA) = (hypotenuse of CBD)/(long side of CBD)
The sides involved here are not corresponding, so this is <em>false</em>.
IV is (long side of CDB)/(hypotenuse of CDB)/(long side of BDA)/(hypotenuse of BDA)
These are corresponding sides, so the proportion is <em>true</em>.
The appropriate choice is ...
... a. I and IV only
