In geometry, it would be always helpful to draw a diagram to illustrate the given problem.
This will also help to identify solutions, or discover missing information.
A figure is drawn for right triangle ABC, right-angled at B.
The altitude is drawn from the right-angled vertex B to the hypotenuse AC, dividing AC into two segments of length x and 4x.
We will be using the first two of the three metric relations of right triangles.
(1) BC^2=CD*CA (similarly, AB^2=AD*AC)
(2) BD^2=CD*DA
(3) CB*BA = BD*AC
Part (A)
From relation (2), we know that
BD^2=CD*DA
substitute values
8^2=x*(4x) => 4x^2=64, x^2=16, x=4
so CD=4, DA=4*4=16 (and AC=16+4=20)
Part (B)
Using relation (1)
AB^2=AD*AC
again, substitute values
AB^2=16*20=320=8^2*5
=>
AB
=sqrt(8^2*5)
=8sqrt(5)
=17.89 (approximately)
Answer:
Step-by-step explanation:
total weight is 4 tons.
ratio of weight of truck to weight of coal is 7:5....added = 12
so 7/12 of the total weight is the truck's
and 5/12 of the total weight is the coal
7/12 of 4 tons = 7/12 * 4 = 28/12 = 2 1/3 tons <== weight of truck
5/12 of 4 tons = 5/12 * 4 = 20/12 = 1 2/3 tons <== weight of coal
Answer:
12 bikes
9 tricycles
Step-by-step explanation:
x+y=21
2x+3y=51
x=-y+21
2(-y+21)+3y=51
-2y+42+3y=51
y+42=51
y=9
x=12
Your answer would be -15.41. If you ever need any other kind of help, just ask me.