Evaluate,the answer should be in the form of an integer.

Mathematics

1 answer:

monitta4 years ago

4 0

∫cos(x)·dx = sin(x)
so your definite integral evaluates to
sin(π/2) - sin(-π/2) = 1 - (-1) = 2

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There are 300 apples and peaches for sale at a farmers market. The ratio of the number of apples to the number of peaches is 7:8

Sidana [21]

Answer:

The ratio of the reaming apples to peaches is 3 : 4

Step-by-step explanation:

* Lets solve the problem

- There are 300 apples and peaches

- The ratio of the number of apples to the number of peaches is 7 : 8

- To find the the numbers of apples and peaches add the terms of the

ratio and then divide the total number of apples and peaches by this

sum and then multiply each terms of the ratio by this quotient

∵ The ratio of apples to peaches = 7 : 8

∴ apple : peach : sum

7 : 8 : 15

? : ? : 300

∴ The number of apples = (300 ÷ 15) × 7 = 20 × 7 = 140 apples

∴ The number of peaches = (300 ÷ 15) × 8 = 20 × 8 = 160 peaches

- 50 apples and 40 peaches are sold

∴ The remaining number of apples = 140 - 50 = 90 apples

∴ The remaining numbers of peaches = 160 - 40 = 120 peaches

- To find the ratio simplify the numbers to its simplest form

∵ There are 90 apples and 120 peaches

∴ apple : peach

90 : 120 ⇒ divided both by 10

∴ 9 : 12 ⇒ divide both by 3

∴ 3 : 4

∴ The ratio of the reaming apples to peaches is 3 : 4

4 0

3 years ago

Find the average value of f over the region D. f(x, y) = 3xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 9)

MakcuM [25]

The average value of f over the region D is 243/4

To answer the question, we need to know what the average value of a function is

<h3>What is the average value of a function?</h3>

The average value of a function f(x) over an interval [a,b] is given by

$\frac{1}{b - a} \int\limits^b_a {f(x)} \, dx$

Now, given that we require the average value of f(x,y) = 3xy over the region D where D is the triangle with vertices (0, 0), (1, 0), and (1, 9).

x is intergrated from x = 0 to 1 and the interval is [0,1] and y is integrated from y = 0 to y = 9

So, $\frac{1}{b - a} \int\limits^b_a {f(x,y)} \, dA = \frac{1}{1 - 0} \int\limits^1_0 \int\limits^9_0 {3xy} \, dxdy \\= \frac{3}{1} \int\limits^1_0 {x} \,dx\int\limits^9_0 {y} \,dy\\ = \frac{3}{1} [\frac{x^{2} }{2} ]^{1}_{0}[\frac{y^{2} }{2} ]^{9}_{0} \\= 3[\frac{1^{2} }{2} - \frac{0^{2}}{2} ] [\frac{9^{2} }{2} - \frac{0^{2}}{2} ] \\= 3[\frac{1}{2} - 0 ][\frac{81}{2} - 0 ]\\= \frac{81}{2} X3 X \frac{1}{2} \\= \frac{243}{4}$