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saveliy_v [14]
3 years ago
10

Hosting a breakfast brunch for 25 people, you plan to serve scrambled eggs.

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
5 0
The answer is a no other option
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#2 Find the value of x. Round to the nearest degree.
Sergio [31]

Answer:

49°

Step-by-step explanation:

arccos x = 15/23

4 0
3 years ago
Can anyone help me please ​
Vikki [24]

Answer:

i think it is x= -4/3

Step-by-step explanation:

maybe

7 0
3 years ago
Read 2 more answers
Can I please get help? Find M1 and M2
marishachu [46]

Answer:

m∠1 = 110° and m∠2 = 70°

Step-by-step explanation:

m∠1 = 110° since vertical angles

To find m∠2, m∠1 and m∠2 must add to 180°

m∠1 + m∠2 = 180°

110° + m∠2 = 180°

m∠2 = 70°

8 0
2 years ago
Suppose you hit a fly ball with an initial upward velocity of 20 feet per second. Which of the following equations would be a re
baherus [9]
A=g≈-32

v=⌠a dt  since a≈-32

v=-32t+C, where C is the initial velocity, which we are told is 20 ft/s

v=-32t+20

h=⌠v dt

h=-32t^2/2+20t+C, where C is the initial height, so what is a reasonable initial height?  How about 3 ft since we are swinging a bat around maybe our waist level... maybe :P

h=-16t^2+20t+3

Your choices may have made a different assumption about the initial height of course but you did not show your choices, but certainly it will be:

h(t)=-16t^2+20t+hi, where hi is the initial height in feet.

This of course ignores air resistance and flight dynamics of a spinning ball with seams, which will make a significant difference in real life :)
3 0
3 years ago
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
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