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Helen [10]
2 years ago
14

How do you find area of a circle

Mathematics
1 answer:
ELEN [110]2 years ago
4 0
A=pir2
Area equals pi(3.16…) times radius squared
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Caold someone PLEASE HELP me with number 56
dusya [7]
What's the question?
5 0
3 years ago
The summit of a mountain pass is 8,050 ft above sea level. the valley below is 12 and 1/2 ft below sea level. what is the vertic
Veseljchak [2.6K]

Answer:

Vertical distance between the valley and the summit = 8062.5 ft

Step-by-step explanation:

Distances above the sea level are denoted by positive notations and below the sea level by negative notation.

Summit of a mountain pass = 8050 (above the sea level)

Valley below the sea level = 12 and half ft = (-12.5) ft

Distance between the summit and the valley = 8050 - (-12.5)

                                                                           = 8050 + 12.5

                                                                           = 8062.5 ft

Therefore, Vertical distance = 8062.5 ft will be the answer.

5 0
3 years ago
Convert into mixed number 81/3 and -29/2
LuckyWell [14K]
Hmmmm to convert from improper to mixed, simply do the division of both, the quotient gets upfront, the remainder as the numerator, for example,

how many times 3 goes into 81?  well, 27 times, with a remainder of 0, thus  \bf 27\frac{0}{3}

how many times does 2 go into 29?  well  14 times, with a remainder of 1, thus  \bf -14\frac{1}{2}
8 0
3 years ago
Read 2 more answers
The following set of coordinates most specifically represents which figure? (−5, 6), (−1, 8), (3, 6), (−1, 4) (6 points) Paralle
Vladimir [108]

Answer:

Rhombus

Step-by-step explanation:

The given points are A(−5, 6), B(−1, 8), C(3, 6), D(−1, 4).

We use the distance formula to find the length of AB.

|AB|=\sqrt{(-1--5)^2+(8-6)^2}

|AB|=\sqrt{16+4}

|AB|=\sqrt{20}

The length of AD is

|AD|=\sqrt{(-1--5)^2+(6-4)^2}

|AD|=\sqrt{16+4}

|AD|=\sqrt{20}

The length of BC is:

|BC|=\sqrt{(-1-3)^2+(8-6)^2}

|BC|=\sqrt{16+4}

|BC|=\sqrt{20}

The length of CD is

|CD|=\sqrt{(-1-3)^2+(6-4)^2}

|CD|=\sqrt{16+4}

|CD|=\sqrt{20}

Since all sides are congruent the quadrilateral could be a rhombus or a square.

Slope of AB=\frac{8-6}{-1--5}=\frac{1}{2}

Slope of BC =\frac{8-6}{-1-3}=-\frac{1}{2}

Since the slopes of the adjacent sides are not negative reciprocals of each other, the quadrilateral cannot be  a square. It is a rhombus

7 0
3 years ago
Read 2 more answers
Two less than a number
loris [4]

Answer:

x-2

Step-by-step explanation:

two less (-2) than a number (variable: x)

7 0
3 years ago
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