A 23 ft ladder is placed against a college classroom building at a point 20 ft above the ground. What is the distance from the b ase of the ladder to the building
1 answer:
The ladder forms the hypotenuse of a right triangle. Solve for a <span>a^2 + b^2 = c^2 </span> <span>a^2 + 9^2 = 18^2 </span> <span>a^2 + 81 = 324 </span> <span>a^2 = 243 </span> <span>a = 15.589 = 15.6 feet</span>
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Find how many hours she will be walking by dividing 4 by 2.5 to get 1.6 hours. Multiply .6 by 60 to see about how many minutes of the hour there will be and get 36. Now add 1 hour and 36 minutes to 6:30 and get 8:06 am
60 divided by 48 equals 1.25 1.25 times 7 equals 8.75 8.75 meters
1/2. Because 3 cancel 3 and 2 can go in 4 twice.
So, P1P2 =
Answer:
$15.679216
Step-by-step explanation:
new sp = 80% of cp
= 0.8 x 199.99
= 159.992
new sp2 = 120% of cp2
= 1.2 x 159.992
÷ 191.9904
new sp3 = 80% of cp3
= 0.8 x 191.9904
= 153.59232
new sp4 = 120% of cp4
= 1.2 x 153.59232
= 184.310784
difference = 199.99 - 184.310784
=15.679216