Answer:
See below ~
Step-by-step explanation:
Given :
⇒ m∠1 = m∠2
⇒ HD = GF
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To Prove :
<u>Δ EHD ≅ Δ EGF</u>
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Solving :
⇒ m∠1 = m∠2 (Given)
⇒ HD = GF (Given)
⇒ ∠E = ∠E (Common angle)
⇒ ΔEHD ≅ ΔEGF (AAS congruence)
Answer:
x = 2.5 or -0.8
Step-by-step explanation:
Here, we are to use completing the square method to solve for the values of x
Firstly, we divide through by 4
= x^2 -7/4x -2 = 0
Now, we move the two term to the right hand side
x^2 -7/4x = 2
Now, we divide the coefficient of x by 2 and square it;
That would be;
(-7/4 * 1/2)^2 = 49/64
we now add this value to both sides of the equation
x^2 -7/4x + 49/64 = 2 + 49/64
The right hand side can be rewritten as;
(x-7/8)^2 = 177/64
taking the square root of both sides
x-7/8 = √(177/64)
x = 7/8 ± √(177/64)
x = 7/8 + √(177/74) or 7/8 - √(177/64)
= x = 2.53802 or -0.788017
Which to the nearest tenth is
x = 2.5 or -0.8
x=3
multiply 2 to both sides to cancel denominator
subtract 8x from both sides and then divide by 2
Answer:
during the first 3mins the bus was accelerating from 0km to 25 kilometers
during the next 90 mins btwn 30mins to 120 mins it was moving at a constant distance n speed
lastly for the last 60 mins from 120 mins to 180 mins it started decelerating from a distance of 25km to 0 km
(5/8):(1/4)= 5/8 • 4= 20/8=5/2 = 2 and 1/2
