This is an isosceles right triangle (AB = BC & ∠ B=90° - Given)
Then the angles at the base are equal and ∠ CAB = ∠ ACB = 45°
Theorem: Segment DE, joining the midpoints of 2 sides is:
1st) parallel to the 3rd side and
2nd) equal to half the measurement of the 3rd side
So if the 3rd side (hypotenuse) = 9 units, DE = 9/2 = 4.5 units
If a square-shaped pool with side lengths are 3a^5 then find
the area of the pool
solution:
A = (3a^5) x (3a^5)
= 9a^5+5
= 9a^10
The area of the square-shaped pool are 9a^10 based on the given side length which is 3a^5
Hey!
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Let's Solve A:
1/2 = 0.5
0.5 + 0.30 = 0.80
a = 0.80
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Let's Solve B:
3/4 = 0.75
0.10 + 0.75 = 0.85
b = 0.85
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Let's Solve C:
1/3 ≈ 0.33
0.33 + 0.50 = 0.83
c = 0.83
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Let's Solve D:
1/3 ≈ 0.33
0.33 + 0.40 = 0.73
d = 0.73
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Answer:
By solving each equation we can see that option A has the lowest value!
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Hope This Helped! Good Luck!
Answer:
1. Group C; 2. Group B; 3. Group D; 4. Group A
Step-by-step explanation:
These equations are in the form
, where v₀ is the initial velocity and h₀ is the initial height.
The first equation has no value for v₀ and a value of 19 for h₀. This means there is no velocity, so the ball is dropped, and since the initial height is 19, it is dropped from 19 meters. This makes it group C.
The second equation has a value of 50 for v₀ and no value for h₀. This means the initial velocity is 50 and there is no initial height. This makes it group B.
The third equation has no value for v₀ and a value of 50 for h₀. This means there is no initial velocity, so the ball is being dropped, and the initial height is 50. This makes it group D.
The fourth equation has a value of 19 for v₀ and no value for h₀. This means the initial velocity is 19 and there is no initial height. This makes it group A.