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kogti [31]
3 years ago
8

joe saved up 25 his mother gave him an additional 20 he boua movie that cost 15 how much money dose joe have left

Mathematics
2 answers:
densk [106]3 years ago
7 0

Answer:

25+20-15=30

Step-by-step explanation:

Here You Go Bud

alina1380 [7]3 years ago
4 0

Answer: 30

Step-by-step explanation: Before Joe had $45 in total, He then spent it on a movie which is $15. So $45-$15=30

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Inez waters her plants every two days. She trims them every 15 days. She did both today. When will she do them both again?
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The answer is 30 days.

The solution for this is to find the least common multiple. By getting the multiple of both numbers.

Multiples of 2:  2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40…..

Multiples of 15: 15,30,45,60,75……..

The Least Common Multiples of 2 and 15 is 30. So, Inez will do them both again in 30 days.

The least common multiple (LCM) of two numbers is the smallest number that is multiple by the both number.

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The reflection of the function f(x) = |x| is -f(x) = -|x|. If we name the reflected function f(x), we have
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4 years ago
Read 2 more answers
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

<em>On integration , we get</em>

=   (\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))

After simplification and cancellation we get

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

on calculation , we get

= \frac{124,416}{5} + \frac{864}{3} + 180

On L.C.M  15

= \frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

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

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