F(2)=8 and f(5v)=25v^2+15v-2
<h3>Answer: A. 5/12, 25/12</h3>
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Work Shown:
12*sin(2pi/5*x)+10 = 16
12*sin(2pi/5*x) = 16-10
12*sin(2pi/5*x) = 6
sin(2pi/5*x) = 6/12
sin(2pi/5*x) = 0.5
2pi/5*x = arcsin(0.5)
2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n
2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n
x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)
x = 5/12+5n or x = 25/12+5n
these equations form the set of all solutions. The n is any integer.
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The two smallest positive solutions occur when n = 0, so,
x = 5/12+5n or x = 25/12+5n
x = 5/12+5*0 or x = 25/12+5*0
x = 5/12 or x = 25/12
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Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.
P / 100 = 25 / 75;
p = 2500 / 75 ;
p = 33.33;
p% = 33.33%
Let the total toys be x.
The first has (1/10)x = x/10
The third child has one more toys than the first = x/10 + 1
The fourth has double of the third: = 2* (x/10 + 1) = 2x/10 + 2
Since we know that the second child has 12 more toys than the first
Therefore the Second minus First = 12.
First let us find the second.
1st + 2nd + 3rd + 4th = x
x/10 + 2nd + (x/10 + 1) + (2x/10 + 2) = x
2nd + x/10 + x/10 + 1 + 2x/10 + 2 = x
2nd + x/10 + x/10 + 2x/10 + 1 + 2 = x
2nd + 4x/10 + 3 = x
2nd = x - 4x/10 - 3
2nd = 6x/10 - 3
But recall
2nd - 1st = 12
6x/10 - 3 - x/10 = 12
6x/10 - x/10 = 12 + 3
5x/10 = 15
5x = 15 * 10
x = 15*10 /5
x = 30
So there were 30 toys in all.
