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svp [43]
4 years ago
11

Please help , 1-10 , thanks !

Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
6 0

Answers:

22

9

20

17

20

35

26

32

23

9

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Calculate the double integral. $$\iint_{R}{\color{red}4} xye^{x^{2}y}\hspace*{3pt}dA, \quad R = [0, 1] \times [0, {\color{red}7}
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Answer:

\mathbf{\int \int _R \ 4xy e^{x^2 \ y}  \ dA =  2 (e^7 -8)}

Step-by-step explanation:

Given that:

\int \int _R 4xye^{x^2 \ y} \ dA, R = [0,1]\times [0,7]

The rectangle R = [0,1] × [0,7]

R = { (x,y): x ∈ [0,1] and y ∈ [0,7] }

R = { (x,y): 0 ≤ x ≤ 1 and 0 ≤ x ≤ 7 }

\int \int _R \ 4xy e^{x^2 \ y}  \ dA = \int^{7}_{0}\int^{1}_{0} 4xye^{x^2 \ y} \ dx dy

\int \int _R \ 4xy e^{x^2 \ y}  \ dA = \int^{7}_{0} \begin {bmatrix} ye^{yx^2} \dfrac{4}{2y} \end {bmatrix}^1 _ 0 \ dy

\int \int _R \ 4xy e^{x^2 \ y}  \ dA = \int^{7}_{0} \begin {bmatrix} ye^{y1^2} \dfrac{4}{2y} - ye^{y0^2} \dfrac{4}{2y} \end  {bmatrix}\ dy

\int \int _R \ 4xy e^{x^2 \ y}  \ dA = \int^{7}_{0} \dfrac{4}{2}(e^y -1) \ dy

\int \int _R \ 4xy e^{x^2 \ y}  \ dA =  \dfrac{4}{2}[e^y -1]^7_0 \ dy

\int \int _R \ 4xy e^{x^2 \ y}  \ dA =  2 [(e^7 -7)-(e^0 -0)]

\int \int _R \ 4xy e^{x^2 \ y}  \ dA =  2 [(e^7 -7)-1]

\mathbf{\int \int _R \ 4xy e^{x^2 \ y}  \ dA =  2 (e^7 -8)}

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4 years ago
A cookie recipe states for every 3 cups of flour, 1 1/2 teaspoons of vanilla are needed. How many teaspoons are needed for 5 cup
34kurt
You would need 2 & 1/2 because if for 3 you need 1 & 1/2 ( which is half) then for 5 you would divide it by 2 and get 2 & 1/2

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I belive the equation would be...

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