Answer:
98 ft²
Step-by-step explanation:
There are a couple of ways you can think about this one. Perhaps easiest is to treat it as a square with a triangle cut out of it. The cutout triangle has a base (across the top) of 14 ft and a height of 14 ft, so its area is ...
A = (1/2)(14 ft)(14 ft) = 98 ft²
Of course the area of the square from which it is cut is ...
A = (14 ft)² = 196 ft²
So, the net area of the two triangles shown is ...
A = (196 ft²) - (98 ft²) = 98 ft²
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Another way to work this problem is to attack it directly. Let the base of the left triangle be x. Then the base of the right triangle is 14-x, and their total area is ...
A = A1 + A2 = (1/2)(x ft)(14 ft) + (1/2)((14-x) ft)(14 ft)
We can factor out 7 ft to get ...
A = (7 ft)(x ft + (14 -x) ft)
A = (7 ft)(14 ft) = 98 ft²
Answer:
5.83 units
Step-by-step explanation:
See attached picture. You can use the Pythagorean Theorem. I added some lines to show the legs of a right triangle. AB is the hypotenuse.
Answer:
I think answer for ur question is 60
To find the average velocity in a velocity-time graph at a particular interval, simply determine the gradient at that particular interval.
<span>a. average velocity= 4/1 </span>
<span>= 4m/s </span>
<span>b. average velocity from 1 to 2.5s= 6/(2.5-1) </span>
<span>= 4m/s </span>
<span>average velocity from 2.5 to 4.0s= 0m/s </span>
<span>average velocity from 0 to 4.0s= (4+0)/4 </span>
<span>= 1m/s </span>
<span>c. average velocity from 1.0 to 4.0s= (4/3)m/s </span>
<span>average velocity from 4.0 to 5.0s= 2/1 </span>
<span>= 2.0m/s </span>
<span>average velocity from 1.0 to 5.0s= ((4/3)+2)/4 </span>
<span>= (5/6)m/s </span>
<span>d. average velocity from 0 to 4.0s= 1.0m/s </span>
<span>average velocity from 4.0 to 5.0s= 2.0m/s </span>
<span>average velocity from 0 to 5.0s= (1.0+2.0)/5 </span>
<span>= (3/5)m/s </span>
The answers are b 41 and c 14 because they are both bigger than 4.0
