3 years ago

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

Mathematics

1 answer:

kipiarov [429]3 years ago

7 0

Answer:

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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3 years ago

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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:

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

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

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

$=18+3b-2b-1$

$\mathrm{Group\:like\:terms}$

$=3b-2b+18-1$

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

$=3b-2b+17$

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

$=b+17$

[RevyBreeze]

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2 years ago

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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

Georgia [21]

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 $nl-mk,ml+nk\in \mathbb{Z}$ .

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3 years ago