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Basile [38]
2 years ago
10

Myra won the hurdles event for the seventh graders. She completed the race in 22.8 seconds and knocked down 1 hurdle. Byron won

the race for the eighth graders. He completed the race in 20.3 seconds and knocked down 3 hurdles. If 2 seconds are added to the finishing time for each hurdle that is knocked down, what is the final time for each runner? Award 110 points to the team with the fastest time.
pls help im tired.
Mathematics
1 answer:
gregori [183]2 years ago
6 0

Answer:

Myra: 24.8s (winner) Byron: 26.3s

Step-by-step explanation:

plz make as crown if someone else answer's

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The cheeseburger in Alan's kid's meal has 300 calories. His fries have 101 and his drink has 150. How many tota calories
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An expression is shown below.<br><br><br> What is the value of the expression?
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Answer:

Step-by-step explanation:

First you need to set up the equation 2/3+1/4/1/2, and we know that there is a stratgey for dividing fractions as well. But first, we need to find the least common fator for 2/3 and 1/4 which is twelve which would be 8/12+3/12 which is 11/12 and 11/12 dividing by 1/2 is 22/12 since all you need to do is do the keep change flip method which would give you an answer of 1 10/12 or 1 5/6

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3 0
3 years ago
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
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