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Q15. Ken earned x dollars at his part-time job on Friday. His wife earned $12 more

mafiozo [28]

Ken earns 54 dollars at his part-time job on Friday

Given that Total amount earned by ken and his wife at their part-time job on Friday= $174

Money earned by ken= x dollars

Money earned by ken's wife = (2x+12) dollars

Total amount earned by ken and his wife at their part-time job on Friday= x dollars + (2x+12) dollars

(x + 2x + 12 ) dollars = $ 174 ( addition of ken and ken's wife income and total amount earned by ken and his wife at their part-time job on Friday is $174)

(3x + 12 ) dollars = $ 174

3x=$162

x= $\frac{162}{3}$

x= 54 dollars

Therefore,54 dollars were earned by ken at his part-time job on Friday

Hence,Ken earns 54 dollars at his part-time job on Friday

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Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$