I’m pretty sure it would be a and b
Step-by-step explanation:
y = 3 + 8x^(³/₂), 0 ≤ x ≤ 1
dy/dx = 12√x
Arc length is:
s = ∫ ds
s = ∫₀¹ √(1 + (dy/dx)²) dx
s = ∫₀¹ √(1 + (12√x)²) dx
s = ∫₀¹ √(1 + 144x) dx
If u = 1 + 144x, then du = 144 dx.
s = 1/144 ∫ √u du
s = 1/144 (⅔ u^(³/₂))
s = 1/216 u^(³/₂)
Substitute back:
s = 1/216 (1 + 144x)^(³/₂)
Evaluate between x=0 and x=1.
s = [1/216 (1 + 144)^(³/₂)] − [1/216 (1 + 0)^(³/₂)]
s = 1/216 (145)^(³/₂) − 1/216
s = (145√145 − 1) / 216
<em>The result can be shown in both exact and decimal forms.
</em>
<em>Exact Form:
</em>
<em>4
e
+
21
</em>
<em>Decimal Form:
</em>
<em>31.87312731
…</em>
Thanks,
<em>Deku ❤</em>
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:
5.25 + 2.50 = 7.75. 10 - 7.75 =2.25
Step-by-step explanation: