<h3>
Answer: 110 square miles</h3>
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Explanation:
There are a few ways to do this.
One method involves drawing a horizontal line to form three rectangles as show below. Note the labels A,B,C.
- Rectangle A in the upper left corner has area of base*height = 6*7 = 42 square miles. I'll let A = 42 since we'll use it later.
- Rectangle B is 20 miles across horizontally (base) and 2 miles tall vertically (height). The 2 miles is from 9-7 = 2. The area of rectangle B is 20*2 = 40 square miles. Let B = 40.
- Then finally, C = 28 because rectangle C is 4 miles across and 7 miles tall, so 4*7 = 28.
Add up those three sub areas found: A+B+C = 42+40+28 = 110 square miles
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There are other methods you could do. A second method is to draw two vertical lines to form 3 other rectangles, then add up the sub areas to get 110.
A third method is to draw a horizontal line across the top to form one large rectangle (20 mi by 9 mi) and subtract off the area of the 7 mi by 10 mi inner rectangle (the empty space), so you'd say 20*9 - 7*10 = 180-70 = 110
Answer:
If you have *dyslexia, then how do you know how to spell so well?
Step-by-step explanation:
Thank you for posting your question here at brainly. The choices based on the above question are the below, it can be found elsewhere.
A. 1/7 <span>mpg
B.
7 mpg
C.
35 mpg
D.
175 mpg
The </span>unit rate for the distance traveled to the amount of gasoline used is 7 mpg. The answer is B.
Answer: you would do lxw and W=5 and L=2 so what u would do it 5x2=10 so your answer is 10
Answer: 3.61×10^5 A
Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.
The formulae is given below as
B = u×Ia²/2(x²+a²)^3/2
Where B = strength of magnetic field on the axis of a circular loop = 4.15T
u = permeability of free space = 1.256×10^-6 mkg/s²A²
I = current on loop =?
a = radius of loop.
Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m
x = distance of the sensor away from center of loop = 2.10 cm = 0.021m
By substituting the parameters into the formulae, we have that
4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2
4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2
4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)
4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I
I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9
I = 4.80×10^-4 / 1.32665 ×10^-9
I = 3.61×10^5 A
