set h=0 and solve for <span>t
</span>So: 0 = 32t - 16^2
<span>32t−16<span>t2</span>=0</span><span>16t(2−t)=0</span><span><span>t=2
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B. is the answer all it asking you is she correct that she only has 18.25 to play with for the games.
Answer:
150
Step-by-step explanation:
1. 4^3 = 4 x 4 x 4 = 16 x 4 = 64
2. 64 - 4 = 60
3. 60 ÷ 2 = 30
4. 30 ⋅ 5 = 150
The answer is lizeth earns 0.35 cents more per hour. beacuse 66 divided by 8 =8.25 then 68.80 divided by 8 is 8.60 then if you subtract 8.60-8.25 it equals 0.35
<span><span>y = 2 + 2sec(2x)
The upper part of the range will be when the secant has the smallest
positive value up to infinity.
The smallest positive value of the secant is 1
So the minimum of the upper part of the range of
y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4
So the upper part of the range is [4, )
The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.
The largest negative value of the secant is -1
So the maximum of the lower part of the range of
y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0
So the lower part of the range is (, 0].
Therefore the range is (, 0] U [4, )
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