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Andrew [12]
3 years ago
12

Given an example of two terms that are like terms and two terms that are not like terms.

Mathematics
2 answers:
kotykmax [81]3 years ago
6 0

Answer:

2x and 5x are like terms, 2x and 3y are not. 3 and 5 are like terms, 6y and 2 are not. etc.

Step-by-step explanation:

For something to be a like term, they either have to have the same variable (ex: x and y and any other variable that may be used) or both be whole numbers (1, 2, 3, etc.)  No mixing the two.

Marat540 [252]3 years ago
3 0

Answer:

example of two like terms:

  • 5xy and 71xy

example of two unlike terms:

  •  4b and 78a

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Can you please tell me what the answer is
Gekata [30.6K]

Answer:

52.5

Step-by-step explanation:

WX curve is 105 degrees

so to find WYX, you need to divide the WX curve by 2:

105 / 2 = 52.5 degrees

I think this is right, I have not done this in years so I am rusty at this

3 0
3 years ago
Determine side x. Round to the nearest unit. 20 9 12 5
tatyana61 [14]

Answer:

12

Step-by-step explanation:

First, you will have to use trigonometry. Since the side of the triangleis the opposite of the angle 53 and the hypotenuse, you will use sin.

Sin 53 degrees = opposite side / hypotenuse side.

The opposite side is x and the hypotenuse side equals 15.

sin53= x/15

15 ( sin 53 ) =x

x= 11.97953

Round

x=12

6 0
3 years ago
Please show me the steps to this thank you.
irakobra [83]
Since the lines are perpendicular the opposite angle of labeled is also 60° and so would the two angles on the the lower just opposite way therefore 8x-4=60
-4 -4
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5 0
3 years ago
Sarah has some chocolates.
kakasveta [241]

Answer:

24:8 which is also just 6:2 or 3:1

Step-by-step explanation:

There are many ways you can write this, and this is what I think hope this helps

8 0
3 years ago
P is inversely proportional to the cube of (q-2) p=6 when q=3 find the value of p when q is 5
sveticcg [70]
\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{  k}}}{x}
\end{array}\\\\
-----------------------------\\\\
\textit{p is inversely proportional to the cube of (q-2)}\implies p=\cfrac{k}{(q-2)^3}
\\\\\\
now \quad 
\begin{cases}
p=6\\
q=3
\end{cases}\implies 6=\cfrac{k}{(3-2)^3}

solve for "k", to find k or the "constant of variation"

then plug k's value back to \bf p=\cfrac{k}{(q-2)^3}

now.... what is "p" when q = 5?  well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
4 0
3 years ago
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