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Bezzdna [24]
3 years ago
13

WHAT WOULD THIS BE? HOW WOULD I DO IT PLEASE HELP IM STUCK

Mathematics
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

90

Step-by-step explanation:

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Hexagon ABCDEF → hexagon A'B'C'D'E'F'. How can the transformation shown be described? Check all that apply.
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The correct answer would be c,and f 
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Which expressions are equivalent to 3(6+b)-(2b + 1)?
Anna [14]

Answer:

18+3b-2b-1

b+17

Step-by-step explanation:

\mathrm{Using\:the\:distributive\:law}:\quad \:-\left(a+b\right)=-a-b

-\left(2b+1\right)=-2b-1

=3\left(6+b\right)-2b-1

=18+3b-2b-1

\mathrm{Group\:like\:terms}

=3b-2b+18-1

\mathrm{Subtract\:the\:numbers:}\:18-1=17

=3b-2b+17

\mathrm{Add\:similar\:terms:}\:3b-2b=b

=b+17

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2 years ago
A local Little league has a total of 60 players, 80% of who are right-handed. Write and solve a proportion that could be used to
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Answer:

48 people are right handed. 12 are left handed

Step-by-step explanation:

80% of 60 is 48.

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3 years ago
A local carpet company has been hired to carpet a planetarium which is in the shape of a circle. If the radius of the planetariu
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Show that if x and y may both be written as the sum of the squares of two rational integers, then their product xy may also be w
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Answer:

See proof below

Step-by-step explanation:

One way to solve this problem is to "add a zero" to complete the required squares in the expression of xy.

Let x=m^2+n^2 and y=l^2+k^2 with m,n,l,k\in \mathbb{Z}. Multiplying the two equations with the distributive law and reordering the result with the commutative law, we get xy=(m^2+n^2)(l^2+k^2)=m^2l^2+m^2k^2+n^2l^2+n^2k^2=n^2l^2+m^2k^2+m^2l^2+n^2k^2

Now, note that 0=2lkmn-2lkmn=2nkml-2nlmk by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain xy=n^2l^2-2nlmk+m^2k^2+m^2l^2+2nkml+n^2k^2=(nl-mk)^2+(ml+nk)^2, thus xy is the sum of the squares of nl-mk,ml+nk\in \mathbb{Z}.

7 0
3 years ago
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