Answer:
see below
Step-by-step explanation:
at least one x-intercept - where it crosses the x axis- They both have at least one x intercept
at least one y-intercept - where it crosses the y axis- they both have at least one
an oblique asymptote or slant asymptote- when the degree of the polynomial in the numerator is higher than the denominator - this is true
a vertical asymptote - they both have a vertical asymptote
the domain - is all real values except x = -3 in one case and x=3 in the other
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!
10 = Ten
100 = One Hundred
1,000 = One Thousand
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10 = 10^1
100 = 10^2 = 10 x 10
1,000 = 10^3 = 10 x 10 x 10
Answer:
-7, -
-
, 9
Step-by-step explanation:
