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3 years ago

I need help with this problem.

Mathematics

2 answers:

Aleksandr [31]3 years ago

8 0

Same y intercept as x+4y=8 through (4,3)

Y intercept is when x=0, so 4y=8, so y=2 and the y intercept is (0,2)

Answer for y intercept: (0,2)

So we need the line through (0,2) and (4,3). Point-point form says the line through (a,b) and (c,d) is

(c-a)(y-b) = (d-b)(x-a)

(4 - 0)(y - 2) = (3 - 2)(x - 0)

4y - 8 = x

Answer for the line: x - 4y = -8

Check:

(0,2) is on the line: 0-4(2) = -8 check

(4,3) is on the line 4 - 4(3) = -8 check

Maksim231197 [3]3 years ago

6 0

Answer: y = -1/4x - 4

Step-by-step explanation:

y = (-1/4)x + 2 (This is the first linear function.)

(4,3) has slope of (-1/4)x or (1/-4)x

4(y2) - 4(y1= the new y -1(x2) -3(x1) = new x which leads to (0,-4) so y-intercept = -4

so all in all, y = (-1/4)x -4 is the answer

Hopefully, you're able to understand this. It's difficult to explain through typed words rather than visually and through a written example.

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Read 2 more answers

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago

I need help with this problem.​

I need help with this problem.