20% of 200 is 40 so 40 students cannot yet drive.
Area of triangle: 1/2(base)(height)
Area of rectangle: length × width
Find the area of the triangles on the sides first using 1/2(base)(height).
Plug 8 for the base and 6 for the height.
1/2(8)(6) = 24 units²
Multiply 24 by 2 for both of the triangles.
(24)(2) = 48 units²
Find the area of the rectangles of the object using length × width.
For the rectangle on top, you can find the width using the Pythagorean Theorem: a² + b² = c².
Plug in 6 and 8 for a and b, and find c (width).
6² + 8² = c²
36 + 64 = 100
Square root both sides; √100 = 10.
The width of the rectangle on top is 10.
Now plug 20 for the length and 10 for the width.
20 × 10 = 200 units²
For the rectangle on the bottom, plug in 20 for the length and 8 for the width.
20 × 8 = 160 units²
For the rectangle on the side, plug in 20 for the length and 6 for the width.
20 × 6 = 120 units²
Add up all the areas ⇒ 48 + 200 + 160 + 120 = 528 units².
I'm so sorry I know the answer but I can't help if it's a test.
<span>Starting position:
Let B start at origin(O) & A is at (-150 ,0) ie 150 km west of the origin . At 3pm , t= 0
A is changing its position at ( 150-35t) km/h
& B is changing its position at 25t km/h
Therefore
AB^2 = OA^2 + OB^2 /// Pythagoras Theorem
= (150-35t)^2 + ( 25t)^2
= 22 500 - 10 500 t + 1225t^2 +625t^2
= 22 500 - 10 500t + 1850t^2
AB = ( 22 500 - 10 500t + 1850t^2)^1/2
d (AB) /dt = 1/2 * ( 22 500 - 10 500t + 1850t^2)^-1/2 * ( - 10 500 + 3700t)
= ( - 5250 + 1850t) / ( 22 500 - 10 500t + 1850t^2)^1/2
At 3pm , t=0
At 7pm , t= 4
So d (AB) /dt = ( -5250 + 1850*4) / ( 22500 - 10500*4 + 1850*4^2)^1/2
= 2150 / 100.5
= 21.4 km/hr
</span>The distance between the ships changing at 7 P.M. is with a speed of 21.4 km/hr.
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