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Alenkinab [10]
3 years ago
8

If all of the diagonals are drawn from a vertex of a quadrilateral, how many triangles are formed?

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

Answer:

4

Step-by-step explanation:

From the chioces give it would be 4

But it is actually 8

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If I have "at least" $20 in my pocket, what are some
Reptile [31]

Answer: $20.01

$20.99

$20

$80

$1,000

Step-by-step explanation:

The reason why is because they are $20 or more.  if it was $19.99 then no.  

5 0
3 years ago
Units of Length Which conversion factor would you use to convert from Meistof 03048 m Customary System Units Metric System Units
melamori03 [73]

Answer:

ight these do be some big number right there

Step-by-step explanation:

3048m+25= 151 * 0.30-167=254

8 0
3 years ago
A 2-inch-wide frame is to be built around theregular decagonal window shown. At what angles a and bshould the corners of each pi
OLEGan [10]
Right angles is one angle.
7 0
4 years ago
How many solutions are there? And what is the answers?
mariarad [96]

Answer:

x = 4 y = 3

Step-by-step explanation:

4 + 3 = 7

4 - 3 = 1

There is only one answer

6 0
3 years ago
Part A: The Sun produces 3.9 ⋅ 1033 ergs of radiant energy per second. How many ergs of radiant energy does the Sun produce in 3
Nuetrik [128]

5914 1404 393

Answer:

  A) 1.3×10^37 ergs

  B) 1.435×10^3 mm

Step-by-step explanation:

A) The amount of energy will be the product of the energy rate and time:

  (3.9×10^33 ergs/s)×(3.25×10^3 s) =12.675×10^(33+3) ergs

  = 1.2675×10^(1+36) ergs

  = 1.2675×10^37 ergs ≈ 1.3×10^37 ergs

The mantissa of the result is the product 3.9×3.25, adjusted to have one digit left of the decimal point. The exponent of the result is the sum of the exponents of the factors, adjusted by 1 to match the adjustment in the mantissa.

The final value should be rounded to 2 significant figures, reflecting the precision of the sun's energy production.

__

B) A millimeter is a small fraction of an inch. 10^-3 mm is a small fraction of the width of a human hair, so 1.435×10^-3 mm is not a reasonable estimate of the distance between railroad tracks.

On the other hand, 1.435×10^3 mm is 1.435 m, almost 56.5 inches. This is a much more reasonable measurement of the distance between railroad rails.

  1.435×10^3 mm is more reasonable

8 0
3 years ago
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