4y + 3 ≤ y + 6
4y + 3 - 3 ≤ y + 6 - 3
4y ≤ y + 3
4y - y ≤ y - y + 3
3y ≤ 3
3y/3 ≤ 3/3
y ≤ 1
So any value of y less than or equal to 1 (so 1 is included in the solution set) satisfies the inequality. C is the correct answer.
2s+14=12
2s=-14+12
2s=-2
s=-2÷2
s=-1
The 15 lbs at $7=$105
So multiple 5x9 as a test and check results. 105-45=60
Here you can see that 60 is a factor of 6 by 10 so you got lucky and got the answer of 10lbs at $6 & 5lbs at $9=15 lbs at $7 average
Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as
<span>(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). </span>
<span>Next, complete the square. </span>
<span>x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² </span>
<span>Also, for the x in the numerator </span>
<span>x = x + 1/2 - 1/2. </span>
<span>So </span>
<span>(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. </span>
<span>Integrate term by term to get </span>
<span>∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C </span>
<span>b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. </span>
<span>∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C </span>
<span>= 2 √(x) ln(x) - √(x) + C. </span>
<span>c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. </span>
<span>∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C </span>
<span>= arctan(e^x) + C.</span>
