Answer:
5 cm
Step-by-step explanation:
The distance a coiled spring stretches varies directly with the amount of weight hanging on the spring.
Distance = D
Weight = W
Hence:
D ∝ W
D = kW
Where k is constant of proportionality
Step 1
We solve for k
If a spring stretches 2 cm when a 30 gram weight is hung from it.
2 = 30k
k = 2/30
k = 1/15
Step 2
how far will it stretch when a 75 gram weight is hung from the spring?
D = kW
We are solving for D
k = 1/15
D = 1/15 × 75
D = 5 cm
The spring would stretch 5cm with a weight of 75 gram.
-2 * 1/4 = -1/2
-1/2 * 1/4 = -1/8
-1/8*8 1/4 = -1/32
Correct choice is the first one.
Answer:
A = $ 3,283.44
A = P + I where
P (principal) = $ 2,950.00
I (interest) = $ 333.44
Step-by-step explanation:
A = P(1 + r/n)^nt
Where:
A = Accrued Amount (principal + interest)
P = Principal Amount
I = Interest Amount
R = Annual Nominal Interest Rate in percent
r = Annual Nominal Interest Rate as a decimal
r = R/100
t = Time Involved in years, 0.5 years is calculated as 6 months, etc.
n = number of compounding periods per unit t; at the END of each period
Answer:
6
Step-by-step explanation:
100% / 8 questions= each question is worth 12.5 points
12.5*6=75
Therefore 6 were answered correctly
Answer:
b = -ax + 4x + 2
Step-by-step explanation:
8x - 3x + 2 - x = ax + b
~Combine like terms
4x + 2 = ax + b
~Subtract ax to both sides
4x - ax + 2 = b
~Rewrite
b = -ax + 4x + 2
Best of Luck!
