All categories

4 years ago

The average density of matter in metagalactic space is 10^−29 grams per cubic cm. Express this number as a decimal.

Mathematics

1 answer:

ohaa [14]4 years ago

3 0

Answer:

0.000 000 000 000 000 000 000 000 000 01 g/cm³

10 fg/km³

Step-by-step explanation:

10^-29 is 1 in the 29th decimal place to the right of the decimal point.

An appropriate set of SI prefixes can be chosen so that the number isn't so difficult to write accurately. Using femtograms per cubic kilometer, we find that the equivalent number is ...

... (10^-29 g/cm³)×(1 fg/(10^-15 g))×(1 cm/(10^-5 km))³ = 10^(-29-(-15)-3·(-5)) fg/km³

... = 10 fg/km³

You might be interested in

What is the square root of 89

kirill [66]

The square root of 89 is 9.43398113. Please award brainliest answer!

3 0

4 years ago

Read 2 more answers

How can you write the expression (28 + 70) as a multiple of a sum of whole numbers with no common factor? please help

grin007 [14]

I hope this helps you

6 0

3 years ago

Read 2 more answers

A road is 3/5 of a mile long. A crew needs to repave 3/4 of the road. How long is the section that needs to be repaved? Write yo

storchak [24]

Answer:

9/20 mile

Step-by-step explanation:

Take (calculate) 3/4 of 3/5 mile:

3 3

---- * ---- = 9/20

4 5

9/20 mile needs to be repaved.

6 0

3 years ago

Find a5: 0, 3, 8, 15, a5, 35, ... a. 20 b. 22 c. 24 d. 25

marishachu [46]

1X1-1: 0
2X2-1: 3
3X3-1: 8
4X4-1: 15
5X5-1: 24
6X6-1: 35
7X7-1: 48
*8X8-1: 63*

6 0

4 years ago

(6 + root27)/(4 - root3) can be written in the form (r + sroot3)/13

Bogdan [553]

Answer:

r = 33 , s = 18

Step-by-step explanation:

Rationalise the denominator by multiplying both numerator and denominator by the conjugate of the denominator.

The conjugate of 4 - $\sqrt{3}$ is 4 + $\sqrt{3}$ , then

$\frac{(6+\sqrt{27})(4+\sqrt{3}) }{(4-\sqrt{3})(4+\sqrt{3}) }$ ← expand numerator/denominator using FOIL

= $\frac{24+6\sqrt{3}+4\sqrt{27}+\sqrt{81} }{16+4\sqrt{3}-4\sqrt{3}-3 }$

= $\frac{24+6\sqrt{3}+4(3\sqrt{3}+9 }{16-3}$

= $\frac{33+6\sqrt{3}+12\sqrt{3} }{13}$

= $\frac{33+18\sqrt{3} }{13}$

with r = 33 and s = 18

5 0

3 years ago