ANSWER
Area of triangle = b*h/2
24*20/2
24*10
The area of this triangle is 240 ft squared
Hope it 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:
The answer is F: 108
Step-by-step explanation:
144 - 108 = 36
108 ÷ 3 = 36
Answer:
26 cm²
Step-by-step explanation:
The area of the rectangle whose dimensions are shown at the right and bottom is ...
(6 cm)(7 cm) = 42 cm²
The figure is smaller than that by the area of the space whose dimensions are shown at the right and in the middle left:
(4 cm)(4 cm) = 16 cm²
The figure area is then the difference ...
42 cm² - 16 cm² = 26 cm²
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<em>Alternate solution</em>
Draw a diagonal line between the lower right inside corner and the lower right outside corner. This divides the figure into two trapezoids.
The trapezoid at lower left has bases 7 and 4 cm, and height 6-4 = 2 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(7 + 4)(2) = 11 . . . . cm²
The trapezoid at upper right has bases 6 cm and 4 cm and height 3 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(6 + 4)(3) = 15 . . . . cm²
Then the area of the figure is the sum of the areas of these trapezoids, so is ...
11 cm² + 15 cm² = 26 cm²
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<em>Comment on other alternate solutions</em>
There are many other ways you can find the area of this figure. It can be divided into rectangles, triangles, or other figures of your choice. The appropriate area formulas should be used, and the resulting partial areas added or subtracted as required.
You can also let a geometry program find the area for you. (It is 26 cm².)
Distance of 10 blocks
Displacement of 2 blocks north