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3 years ago

How do you answer this?

Mathematics

1 answer:

Cloud [144]3 years ago

8 0

For 19, you would do 6b+8b=175, 14b=175, b=12.5. YZ would be 8b=YZ, 8*12.5=YZ, 100=YZ.

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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

11 months ago

1) Find the slope and the y-byintercept of the line that passes through the points (1,5) and (3,9)

dimulka [17.4K]

Answer: slope= 2, y-intercept= 3

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2 years ago

The parabola y=x^2 is shifted up by 8 units. What is the new equation of the parabola?

elixir [45]

Answer:

y=x^2+8

Step-by-step explanation:

5 0

2 years ago

On Halloween, a man presents a child with a bowl containing eight different pieces of candy. He tells her that she may have thre

likoan [24]

Answer:

$56$ choices

Step-by-step explanation:

We know that we'll have to solve this problem with a permutation or a combination, but which one do we use? The answer is a combination because the order in which the child picks the candy <u><em>does not</em></u> matter.

To further demonstrate this, imagine I have 4 pieces of candy labeled A, B, C, and D. I could choose A, then C, then B or I could choose C, then B, then A, but in the end, I still have the same pieces, regardless of what order I pick them in. I hope that helps to understand why this problem will be solved with a combination.

Anyways, back to the solving! Remember that the combination formula is

$_nC_r=\frac{n!}{r!(n-r)!}$ , where n is the number of objects in the sample (the number of objects you choose from) and r is the number of objects that are to be chosen.