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Vinil7 [7]
1 year ago
10

Is this figure a quadrilateral?

Mathematics
2 answers:
choli [55]1 year ago
8 0

Answer:

Yes its a quadrilateral

Step-by-step explanation:

It has 4 sides

yulyashka [42]1 year ago
6 0

Answer:

yes

Step-by-step explanation:

It is an irregular quadrilateral

Hope this helps

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HOW is it helpful to write numbers in different ways?
Alisiya [41]
It is helpful to write them in different ways
some ways to write them are
1.fractions
2.decimals
3.squareroots (if applicable)

so it would be more helpful in an equation to leave 6/7 in fractional form if  you are going to manipulate it more, because 0.857142857142... is much harder to keep track of than 6/7


and sometimes, they want a percent which is easier to convert to from decimal form than from fractinoal form so ex  0.857142857142...=85.7% vs 6/7 to percent

sometimes there will be square roots and they are easier if left like that ex
√2=1.4142135623...
it would be easier to leav it in square root

it depends on the equation you are trying to solve, because different forms have different pros and cons, some are easier to work with in a certain form but not in another, sometimes, you will need to change between multipule forms during the same problem
8 0
3 years ago
Read 2 more answers
Question 1 of 10
Natali5045456 [20]
I believe it’s letter c if it’s saying 2/4 and reduce down to 1/2
6 0
2 years ago
When comparing two box-plots that show the same type of information, what determines agreement within the data?
Mice21 [21]

Answer:

c.the mean of each data set

5 0
2 years ago
Use a table of values with at least 5 values to graph the following function:
34kurt

The given expression :

y=(\frac{1}{2})^x

For coordinates:

put x = 0 then :

\begin{gathered} y=(\frac{1}{2})^0 \\ y=1 \end{gathered}

Coordinate : (x, y) = (0, 1)

Put x= 1 and simplify :

\begin{gathered} y=(\frac{1}{2})^1 \\ y=\frac{1}{2} \\ y=0.5 \end{gathered}

Coordinate : (x, y) = ( 1, 0.5)

Put x = (-2) and simplify :

\begin{gathered} y=(\frac{1}{2})^{-2} \\ y=\frac{1^{-2}}{2^{-2}} \\ y=\frac{2^2}{1^2} \\ y=2^2 \\ y=4 \end{gathered}

Coordinate : (x, y) = ( -2, 4)

Put x = (-3) and simplify :

\begin{gathered} y=(\frac{1}{2})^{-3} \\ y=\frac{1^{-3}}{2^{-3}} \\ y=\frac{2^3}{1^3} \\ y=2^3 \\ y=8 \end{gathered}

Coordinate : (x, y) = (-3, 8)

Substitute x = (-1) and simplify :

\begin{gathered} y=(\frac{1}{2})^x \\ y=(\frac{1}{2})^{-1} \\ y=\frac{1^{-1}}{2^{-1}} \\ y=\frac{2}{1} \\ y=2 \end{gathered}

Coordinate : (x, y) = ( -1, 2)

So, the coordinates are :

The graph is :

4 0
1 year ago
10p = 66 what is the value of p
Zepler [3.9K]
The value of p is 11. To find this, you must isolate the variable, p, so you divide 10 by both sides of the equation to get p=11.
6 0
3 years ago
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