<h3>The length of longer piece is 30 inches</h3>
<em><u>Solution:</u></em>
Let "x" be the length of longer piece
Let "y" be the length of shorter piece
<em><u>The length of a board is 55 inches</u></em>
Therefore,
x + y = 55 -------- eqn 1
<em><u>The board needs to be cut into 2 pieces with one piece 5 inches longer than the other piece</u></em>
Therefore,
x = 5 + y ------- eqn 2
Substitute eqn 2 in eqn 1
5 + y + y = 55
2y = 55 - 5
2y = 50
<h3>y = 25</h3>
Substitute y = 25 in eqn 1
x + 25 = 55
x = 55 - 25
<h3>x = 30</h3>
Thus the length of longer piece is 30 inches
Answer:
Explanation:
The table that shows the pattern for this question is:
Time (year) Population
0 40
1 62
2 96
3 149
4 231
A growing exponentially pattern may be modeled by a function of the form P(x) = P₀(r)ˣ.
Where P₀ represents the initial population (year = 0), r represents the multiplicative growing rate, and P(x0 represents the population at the year x.
Thus you must find both P₀ and r.
<u>1) P₀ </u>
Using the first term of the sequence (0, 40) you get:
P(0) = 40 = P₀ (r)⁰ = P₀ (1) = P₀
Then, P₀ = 40
<u> 2) r</u>
Take two consecutive terms of the sequence:
- P(1) / P(0) = 40r / 40 = 62/40
You can verify that, for any other two consecutive terms you get the same result: 96/62 ≈ 149/96 ≈ 231/149 ≈ 1.55
<u>3) Model</u>
Thus, your model is P(x) = 40(1.55)ˣ
<u> 4) Population of moose after 12 years</u>
- P(12) = 40 (1.55)¹² ≈ 7,692.019 ≈ 7,692, which is round to the nearest whole number.
For this problem, you're going to need to use trig. More specifically, you need to use the tangent function:
tan (value) = opposite/adjacent
tan 75 = opposite/8 (Here, the opposite side is the length of the ladder)
8 * tan 75 = opposite
29.8564065 = opposite
So, the ladder is 29.9 feet.
y = ln x , 1 <= x <= 3, about x axis and n = 10, dy/dx = 1/ x
S = (b a) ∫ 2π y √( 1 + (dy/dx) ^2) dx
so our f(x) is 2π y √( 1 + (dy/dx) ^2)
(b - a) / n = / 3 = (3-1) / 30 = 1/15
x0 = 1 , x1 = 1.2, x2 = 1.4, x3 = 1.6 ....... x(10) = 3
So we have , using Simpsons rule:-
S10 = (1/15) ( f(x0) + 4 f(x1) + 2 f)x2) +.... + f(x10) )
= (1/15) f(1) + f(3) + 4(f(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)) + 2(f(1.4) + f(1.8) + f(2.2) + f(2.6) )
( Note f(1) = 2 * π * ln 1 * √(1 + (1/1)^2) = 0 and f(3) = 2π ln3√(1+(1/3^2) = 7,276)
so we have S(10)
= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +
