Hello from MrBillDoesMath!
Answer:
(3/4) a^(-5)b^(-3)c^2
Discussion:
(18 a^-3b^2c^6)/ (24 a^2b^5c^4) =
(18/24) a^ (-3-2) b^(2-5) c^(6-4) =
as a^-3/a^-2 = a ^ (-3-2) = a^(-5), for examples
(3/4) a^(-5)b^(-3)c^2
Thank you,
MrB
We'll use standard labeling of right triangle ABC, C=90 degrees, legs a, b, hypotenuse c.
11.
Right triangle, cliff peak A, boat B, angle opposite cliff is B=28.9 deg. adjacent leg a=65.7 m, cliff height is leg b.
tan B = b/a
b = a tan B = 65.7 tan 28.9° = 36.3 m
12.
Similar story, boat at B, opposite b=3.5 m, rope c=12 m
sin B = b/c
B = arcsin b/c = arcsin (3.5/12) = 17.0°
13.
c=124 m, A=58°
sin A = a/c
a = c sin A = 124 sin 58 = 105.2 m
14.
That's a hypotenuse c=4-1.2 = 2.8 m to a height b=1.8m so
cos A = b/c
A = arccos b/c = arccos (1.8/2.8) = 50.0°
15.
Not a right triangle, an isosceles triangle. Half of it is a right triangle with hypotenuse one arm, c=9.8 cm and angle opposite half the base of B=62/2=31°. We're after d=2b:
sin B = b/c
b = c sin B
d = 2b = 2 c sin B = 2(9.8) sin 31 = 10.1 cm
Almost equilateral
Answer: 
Step-by-step explanation:
Let be "x" the time in minutes a 150-pound person must walk at 4 mph to use at least 190 calories.
The amount of calories that a 150-pound person uses in 1 minute when walking at a speed of 4 mph is:
Therefore, knowing this, we can write the following proportion:
Finally, we must solve for "x" in order to find its value.
Multiplying both sides of the equation by 190, we get this result:
From one vertex of an octagon you can draw 5 diagonals.
There are 8 vertices in an octagon, and we are choosing one as our starting vertex. There are then 7 vertices left to draw a line to, but 2 of the vertices are already connected to our main vertex (because they are connected along the side of the octagon). That leaves 5 vertices to draw a diagonal to from our original vertex.
C.) 27 and a half. 12 minutes times 5 equals an hour so 5 and a half pages times 5 equals 27 and a half pages per hour.
