The answer would be 10.81665383 so the nearest whole number would be 11
<span>sin(angle)=<span><span>opposite leg/</span><span>hypotenuse</span></span></span>
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<span><span><span> sin(20) = 10/ hypotenuse</span></span></span>
<span><span><span>hypotenuse = 10/sin(20) = 29.238 ( round off as necessary)</span></span></span>
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Answer:
<h3>Pudge has 12 morethan apples that is 24 apples and both Ace and Christi has not more than 12 apples together they has atmost 12 apples.The apples does each having is</h3><h3>P = Pudge's Apples
=24</h3><h3>A = Ace's Apples
=8</h3><h3>and C = Christi's Apples=4.</h3>
Step-by-step explanation:
Let P = Pudge's Apples
Let A = Ace's Apples
Let C = Christi's Apples
<h3>To find how many apples does each have if Pudge has 12 more than both Ace and Christi together:</h3>
Given that P = 3A , A = 2C and P = A + C + 12
Substitute the value for P in P = A + C + 12 we get
3A = A + C + 12
3A-A=C+12
2A=C+12
From A = 2C we have that 
Substitute the value C:






<h3>∴ A=8</h3>
Substituting the value of A in P=3A we get
P = 3(8)
<h3>∴ P = 24
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Substituting the values of P and A in P = A + C + 12



4=C
Rewritting we get
<h3>∴ C=4 </h3><h3>Hence Pudge has 12 morethan apples that is 24 apples </h3><h3>and both Ace and Christi has not more than 12 apples together they has atmost 12 apples</h3>
Answer:
3/25
Step-by-step explanation:
3 red marbles, 4 green marbles, and 3 blue marbles = 10 marbles
P(green) = number of green / total = 4/10 = 2/5
We replace it
3 red marbles, 4 green marbles, and 3 blue marbles = 10 marbles
P(red) = number of red / total = 3/10
P(green, replace,red) =2/5*3/10 = 3/25
<span>Let a_0 = 100, the first payment. Every subsequent payment is the prior payment, times 1.1. In order to represent that, let a_n be the term in question. The term before it is a_n-1. So a_n = 1.1 * a_n-1. This means that a_19 = 1.1*a_18, a_18 = 1.1*a_17, etc. To find the sum of your first 20 payments, this sum is equal to a_0+a_1+a_2+...+a_19. a_1 = 1.1*a_0, so a_2 = 1.1*(1.1*a_0) = (1.1)^2 * a_0, a_3 = 1.1*a_2 = (1.1)^3*a_3, and so on. So the sum can be reduced to S = a_0 * (1+ 1.1 + 1.1^2 + 1.1^3 + ... + 1.1^19) which is approximately $5727.50</span>