8 can go into 12 only once.
Answer: 
<u>Step-by-step explanation:</u>
I'm going to separate this into sections so it makes more sense for you to read. For the problems with π where you have to round, ask your teacher where to round, unless your textbook specifies it:
A – 100 cm^2
To calculate area of squares, you multiply l • w. It's a square, so all sides are equal, and since we know that one side = 10 cm, the area is 10 • 10 = 100
B – πr^2 (not sure if the r shows up very well, so I'm retyping it in words - pi • radius squared)
C – 25π cm^2 or an approximate round like 78.54 cm^2 (ask your teacher about this – it could be to the nearest tenth, hundredth, etc.)
To find the area of a circle, you must follow the formula πr^2. In this case, the diameter is 10. The radius is half the diameter, so to substitute the values you must find 10 ÷ 2 = 5. So the radius is 5 cm. From there you can substitute r for 5, ending up with π • 5^2. 5^2 = 25, so the area is 25π, or about 78.54, depending on where the question wants you to round.
D – An approximate round (to the nearest hundredth it is 21.46 cm^2)
To find the area of the shaded region, just subtract the circle's area from the square's area, or 100 – 25π ≈ 21.46. Again, though, ask your teacher about where to round, unless your textbook specifies it.
E – dπ (diameter • pi)
F – 10π cm^2 or an approximate round like 31.42 cm^2
The diameter is 10. 10π ≈ 31.42
Hope this helps!
Answer:
BD = 35
Step-by-step explanation:
Calculate CD in right triangle ABC, then BD in right triangle BCD
Using Pythagoras' identity in both triangles.
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
In Δ ADC
CD² + AD² = AC² , substitute values
CD² + 9² = 15²
CD² + 81 = 225 ( subtract 81 from both sides )
CD = 144 ( take the square root of both sides )
CD = 
= 12
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In Δ BCD
BD² + CD² = BC² , substitute values
BD² + 12² = 37²
BD² + 144 = 1369 ( subtract 144 from both sides )
BD² = 1225 ( take the square root of both sides )
BD = 
= 35
