Answer:
Here you go
Step-by-step explanation:
15 points only, bro?
The value of W when V = 9 is 405
<h3>How to solve for W?</h3>
The variation is a direct variation from W to V^2.
This means that:
W = kV^2
Where k is the constant of variation.
W = 320 when V = 8 means that:
320 = k * 8^2
Evaluate
320 = 64k
Divide both sides by 64
k = 5
Substitute k = 5 in W = kV^2
W = 5V^2
When V = 9, we have:
W = 5 * 9^2
Evaluate
W = 405
Hence, the value of W when V = 9 is 405
Read more about direct variation at:
brainly.com/question/6499629
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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!
In fact, both Amanda's and Stephen's profs are correct; they are just using different supplementary angles. Amanda's is using the supplementary angles <span>∠1 and ∠4, and</span> ∠3 and ∠4, whereas Stephen is using ∠1 and ∠2, and <span>∠2 and ∠3.</span><span> </span>Please check the picture to visualize this more effectively.
