Answer:
A: (0,0), B: (3, -4), 5 units
Step-by-step explanation:
A: (0,0), B: (3, -4)
To find the distance, use the distance formula. Square-root((0-3)^2+(0-(-4))^2) ---> Square-root((-3)^2 + 4^2) ---> Square-root(9+16) ---> Square-root(25) ---> 5units.
Answer:
Step-by-step explanation:
{ - 9, - 4, - 1}
Answer:
see below
Step-by-step explanation:
<h3>Given</h3>
- Distance is 142.2 m, correct to 1 decimal place
- Time is 7 seconds, correct to nearest second
<h3>To find:</h3>
- Upper bound for the speed
<h3>Solution </h3>
<em>Upper bound for the speed = upper bound for distance/lower bound for time</em>
- Upper bound for distance = 142.25 m (added 0.1/5 = 0.05)
- Lower bound for time = 6.5 seconds (subtracted 1/2 = 0.5)
<u>Then, the speed is:</u>
- 142.25/6.5 = 21.88 m/s
- 21.88 = 21.9 m/s correct to 1 decimal place
- 21.88 = 22 m/s correct to nearest m/s
Answer:
(B)
Step-by-step explanation:
PRINCIPAL AMOUNT(p) = $400
RATE (r)= 250% = 2.5
INTEREST = $80
t = time (in days) = t/365
By using the formula,
r = 
2.5 = 
t = 
t = 29.2
t = 29 days (approx)
Hence option (B) is correct.
Given :
The diagonals of rhombus ABCD intersect at E.
∠CAD = 20°.
To Find :
The angle ∠CDA.
Solution :
We know, diagonals of a rhombus bisects each other perpendicularly.
So, ∠DEA = 90°.
In triangle ΔEAD :
∠EAD + ∠AED + ∠EDA = 180°
20° + 90° + ∠EDA = 180°
∠EDA = 70°
Now, we know diagonal of rhombus also bisect the angle between two sides .
So, ∠CDA = 2∠EDA
∠CDA = 2×70°
∠CDA =140°
Therefore, ∠CDA is 140°.
