Desmos.com and plug in the equation
Answer:
246.3%
the complete question is found in the attached document
Step-by-step explanation:
1st step:
using 1936 data,
w1= 356% = 3.56(356/100) , H= 79 feet
specific gravity = 2.65
Sₓ= 100%= 1
initial void ratio(e₀)= (w1 x specific gravity)/Sₓ
=3.56 x 2.65/1 = 9.434
2nd step
using 1996 data
ΔH= 22ft
ΔH/H = Δe/(1 + e₀)
22/79 = Δe/(1+9.434)
0.278=Δe/10.434
Δe= 0.278 x 10.434
Δe= 2.905
Δe= e₀ - eₓ
eₓ= e₀-Δe
eₓ= 9.434 - 2.905
eₓ= 6.529
3rd step
calculating water content in 1996
eₓ =6.529, specific gravity= 2.65, Sₓ= 100%
W2 X 2.65 = 1 x 6.529
w2 = 6.529/2.65 = 2.463 = 246.3%
Answer:
B.)
Step-by-step explanation:
use the slope rule
Fill in values
Simplify
Simplify further
Done.
Answer:
Clyde is 49.74 away from the harbor
Step-by-step explanation:
Here in this question, we are interested in knowing the distance of Clyde from the harbor.
The key to answering this question is having a correct diagrammatic representation. Please check attachment for this.
We can see we have the formation of a right angled triangle with the distance between Clyde’s ship and the harbor the hypotenuse.
To calculate the distance between the two, we shall employ the use of Pythagoras’ theorem which states that the square of the hypotenuse is equal the sum of the squares of the two other sides.
Let’s call the distance we want to calculate h.
Mathematically;
h^2 = 25^2 + 43^2
h^2 = 625 + 1849
h^2 = 2474
h = √2474
h = 49.74 miles
The sum of point A and point D is zero
-5+5= 0