Answer:
78
Step-by-step explanation:
in a quadrilateral that's inscribed in a circle, opposite angles are supplementary.
102+78=180
Answer:
the asymptote would be zero because the graph is approaching zero, but never actually touches it.
Answer:
2p+3>2(p−3)
Use the distributive property to multiply 2 by p−3.
2p+3>2p−6
Subtract 2p from both sides.
2p+3−2p>−6
Combine 2p and −2p to get 0.
-3>6
This is true for any p.
p∈R
Step-by-step explanation:
Hope this helps! :)
Answer:
x₂ = 7.9156
Step-by-step explanation:
Given the function ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)
If f(x) = ln(x)+x-10
f'(x) = 1/x + 1
f(9) = ln9+9-10
f(9) = ln9- 1
f(9) = 2.1972 - 1
f(9) = 1.1972
f'(9) = 1/9 + 1
f'(9) = 10/9
f'(9) = 1.1111
x₁ = x₀ - f(x₀)/f'(x₀)
x₁ = 9 - 1.1972/1.1111
x₁ = 9 - 1.0775
x₁ = 7.9225
x₂ = x₁ - f(x₁)/f'(x₁)
x₂ = 7.9225 - f(7.9225)/f'(7.9225)
f(7.9225) = ln7.9225 + 7.9225 -10
f(7.9225) = 2.0697 + 7.9225 -10
f(7.9225) = 0.0078
f'(7.9225) = 1/7.9225 + 1
f'(7.9225) = 0.1262+1
f'(7.9225) = 1.1262
x₂ = 7.9225 - 0.0078/1.1262
x₂ = 7.9225 - 0.006926
x₂ = 7.9156
<em>Hence the approximate value of x₂ is 7.9156</em>
Answer:
D. 2058.24 in³
Step-by-step explanation:
Let r,h be the radius and height of smaller can and R,H be the radius and height of larger can
Then, R=4r and H=4h
Volume of smaller can=32.16 in³
⇒ πr²h=32.16 in³
Volume of larger can= πR²H
= π(4r²)×4h
= 64πr²h
= 64×32.16 in³
= 2058.24 in³
Hence, correct option is:
D. 2058.24 in³
