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lorasvet [3.4K]
4 years ago
9

FIND THE CORRECT INTERCEPT OF THE LINE!

Mathematics
1 answer:
Ber [7]4 years ago
7 0

Answer:

see below

Step-by-step explanation:

The x intercept is where it crosses the x axis

It crosses at x = -7.5

(-7.5,0)

The y intercept is where it crosses the y axis

It crosses at y = 5.5

(0,5.5)

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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}
Lady_Fox [76]

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
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beks73 [17]

Answer

87 measures an acute angle

Step-by-step explanation:

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3 years ago
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Answer:

Step-by-step explanation:

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4 0
3 years ago
Pls sovle thankyou number 7
Aleks [24]

Answer:

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Step-by-step explanation:

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Read 2 more answers
The asymptote of the function f(x) = 3x + 1 – 2 is ___. Its y-intercept is ___.
Svetradugi [14.3K]
The only way this could have an asymptote, either horizontal, vertical, or oblique is if it is a rational function, proper or improper. This function is linear:
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8 0
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