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Fittoniya [83]
3 years ago
11

Triangle ABC was translated according to the rule

Mathematics
2 answers:
crimeas [40]3 years ago
6 0

Answer:

its C ( -5.5, 3.2) i just took the test

Step-by-step explanation:

valkas [14]3 years ago
5 0

Answer:

C). (-5.5, 3.2)

Step-by-step explanation:

I just took the quiz on Edge...  Hope this helps!!

Also heart and rate if you found this answer helpful!!

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Which expression is the same as 10÷4?<br><br> 110 of 4<br><br> 4÷10<br><br> 110×14<br><br> 14 of 10
erica [24]

Answer:

1/4 of 10

Step-by-step explanation:

7 0
3 years ago
Find an expression for the nth term -3 , 2 , 7, 12​
slavikrds [6]

Answer:

5n-8

Step-by-step explanation:

Because this is an arithmetic progression, there is simply an addend that is being added to every number. Because that is 5, the expression, for now, would be 5n. However, the progression does not start at 0, so we will have to subtract 3, and then it would be 5n - 3. We're still not at the answer; this is because the first term starts at 1, not 0. We will have to account for that, so subtract the coefficient, or 5, from the whole expression. That would be 5n - 8, and that would be the answer.

4 0
3 years ago
Find the surface area of the cylinder represented by the trunk of the tree.
Montano1993 [528]

Answer:

43.98

Step-by-step explanation:

8 0
3 years ago
Round 48.078 to the nearest tenth<br><br> 48.1,<br><br> 48.08,<br><br> 50,<br><br> or,<br><br> 48.8
musickatia [10]

your  answer is 48.1

7 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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