Students sold books to the local library as a part of a school drive. paper books were brought for $1.45 each. The library spent
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<span>For the plane, we have z = 5x + 9y
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
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For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
Step-by-step explanation:
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The LCM of 18 and 24 using the prime-factorization method is <u>72</u>.
The LCM of 27, 81, and 54, using the division method is <u>162</u>.
In the first question, we are asked to find the LCM of the given numbers using the prime factorization method.
(i) 18, 24.
Prime factorization gives:
18 = 2*3², and 24 = 2³*3.
The LCM is the product of the highest power of each prime factor.
Thus, LCM = 2³*3² = 8*9 = 72.
Thus, the LCM of 18 and 24 using the prime-factorization method is <u>72</u>.
Doing similarly for other sub-parts, The LCM of:
(ii) 16, 40 is 80.
(iii) 30, 36 is 180.
(iv) 28, 44 is 308.
(v) 20, 32 is 160.
(vi) 20, 135 is 540.
(vii) 45, 75 is 225.
(viii) 36, 84 is 252.
(ix) 12, 18, 24 is 72.
(x) 25, 35, 45 is 1575.
(xi) 9, 15, 21 is 315.
(xii) 25, 50, 75 is 150.
In the second question, we are asked to find the LCM of the given numbers, using the division method.
(i) 27, 81, 54.
The division method gives:
The LCM is the product of all the numbers used for division.
LCM = 2*3*3*3*3*1 = 162.
Thus, the LCM of 27, 81, and 54, using the division method is <u>162</u>.
Doing similarly for other sub-parts, The LCM of:
(ii) 18, 45, 63 is 630.
(iii) 35, 55, 100 is 7700.
(iv) 210, 140, 315 is 1260.
(v) 112, 120, 150 is 8400.
(vi) 144, 180, 300 is 3600.
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