Y = -7sin(x) + 2cos(x)
y = -7sin(3π/4) + 2cos(3π/4)
y = -7sin(9.42/4) + 2cos(9.42/4)
y = -7sin(2.355) + 2cos(2.355)
y = -7(0.03698381721) + 2(0.9993158646)
y = -0.2588867205 + 1.998631729
y = 1.739745009
Answer:
Distance swam by Sonya in 3 weeks = 19740 m
Step-by-step explanation:
To practice for a competition, Sonya swam 0.94 kilometer in the pool each day for 3 weeks.
Number of days in 1 week = 7
Number of weeks she swam = 3
Total number of days she swam = 3 x 7 = 21 days
Distance swam per day = 0.94 km
Total distance swam = Total number of days she swam x Distance swam per day
Total distance swam = 21 x 0.94 = 19.74 km = 19.74 x 1000 m = 19740 m
Distance swam by Sonya in 3 weeks = 19740 m
When x is zero there will never be an interception of the line as x=0 is an asymtope
Answer: 37 units
Step-by-step explanation:
This also works as the height of the triangle.
This also works as the base of the triangle.
Let's call pink ''a'', and blue ''b''. The side we're looking for ''c'' is the hypothenuse.
To find the values of a and b, use the area formula of a square and solve for a side. In this case, since we're going to need the squared values, this step can be omitted.
![s=\sqrt[]{A}](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7BA%7D)
Let's work with Blue.
![s=\sqrt[]{144units^2} \\s=12units](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7B144units%5E2%7D%20%5C%5Cs%3D12units)
Now Pink.
![s=\sqrt[]{1225units^2}\\s=35units](https://tex.z-dn.net/?f=s%3D%5Csqrt%5B%5D%7B1225units%5E2%7D%5C%5Cs%3D35units)
So we have a triangle with a base of 35 units and a height of 12 units.
Now let's use the pythagoream's theorem to solve.
![c^2=a^2+b^2\\c=\sqrt[]{a^2+b^2} \\c=\sqrt[]{(12units)^2+(35units)^2}\\c=\sqrt[]{144units^2+1225units^2}\\ c=\sqrt[]{1369units^2}\\ c=37units](https://tex.z-dn.net/?f=c%5E2%3Da%5E2%2Bb%5E2%5C%5Cc%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%20%5C%5Cc%3D%5Csqrt%5B%5D%7B%2812units%29%5E2%2B%2835units%29%5E2%7D%5C%5Cc%3D%5Csqrt%5B%5D%7B144units%5E2%2B1225units%5E2%7D%5C%5C%20c%3D%5Csqrt%5B%5D%7B1369units%5E2%7D%5C%5C%20c%3D37units)
Answer:
h(8q²-2q) = 56q² -10q
k(2q²+3q) = 16q² +31q
Step-by-step explanation:
1. Replace x in the function definition with the function's argument, then simplify.
h(x) = 7x +4q
h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q
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2. Same as the first problem.
k(x) = 8x +7q
k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q
_____
Comment on the problem
In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...
h(8q² -2q) = 56q² -14q +4z
and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.
In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.
