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Mathematics

2 answers:

TEA [102]3 years ago

5 0

So you would do:

$20.00 and multiply 80%

so 20 x 80% = $16

so your answer would be $16.00

Good Luck!

Kay [80]3 years ago

4 0

$16.00 hope this helps

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vertices of parallelogram ABCD are A(2, 0), B(4, 3), C(3, 1). How many possibilities are there for the position of the fourth ve

nordsb [41]

Answer:

1 possibility is there for the the position of the fourth vertex.

7 0

3 years ago

In a theater, there are 4 long rows of chairs,each with an equal number of chairs There are 5 ahort rows of chairs,each with an

Hoochie [10]

There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.

So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y

When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)

When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)

Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23

Thus we can say the number of chairs increase by 23 as compared to a normal day.

7 0

3 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]

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

On integration , we get

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

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

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