Answer:
h = 15
Step-by-step explanation:
Use Pythagorean Theorem to solve for h.
a² + b² = c²
8² + h² = 17²
64 + h² = 289
h² = 225
h = 15
This is just about similar triangle
mn=2jl=22
mo=2lk=26
jk=2on=40
Answer:
The Figure for Right triangle is below,
Therefore , 15 unit length represents BC.
Step-by-step explanation:
Given:
Consider a right triangle ABC, Such that
To Find:
BC = ?
Solution:
In Right Angle Triangle ABC, Cosine and Tangent identity
BUT,
....Given
On Comparing,
Adjacent side to angle A = AB = 15
Opposite side to angle A = BC = 8
Hypotenuse = AC =17
Also Pythagoras theorem is Satisfies,
The Figure for Right triangle is below,
Therefore , 15 unit length represents BC.
Answer:x=20/y
Step-by-step explanation:
x α 1/y
removing the constant of proportionality sign α and replace it with =k
x = k/y
x=10 y=2
10 = k/2
Cross multiply
10x2=k
20=k
k=20
Substitute k=20 in x=k/y
x=20/y.........required equation
Answer:
A) g(t) = - 16*t² + 80*t
B) domain is ( 0 ≤ t ≤ 5 )
Step-by-step explanation:
A) for the model rocket
h = 0 and v = 80 feet/sec
Then the equation g(t) = -16*t² + v*t + h
became
g(t) = - 16*t² + 80*t + 0 g(t) = - 16*t² + 80*t
B) The equation g(t) = - 16*t² + 80*t is defined for all real numbers then the domain interval for t is ( -∞ , ∞ ). Now in our case, a model rocket was launched is not possible to get negative values for g(t) then the biggest value for t = 5 from
-16*t² + 80*t ≥ 0
-16*t ≥ -80 16*t ≤ 80
t ≤ 5
And the domain is ( 0 ≤ t ≤ 5 )
