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soldi70 [24.7K]
4 years ago
8

Physicists estimate that there are between 1078 and 1082 atoms in the universe. How do these two numbers compare with one anothe

r? If the larger number is right, how many universes with the smaller number of atoms would fit inside the larger universe?
Mathematics
1 answer:
S_A_V [24]4 years ago
4 0

Answer:

10,000 universes with the smaller number of atoms would fit inside the larger universe

Step-by-step explanation:

The correct question is as follows;

Physicists estimate that there are between 10^78 and 10^82 atoms in the universe. How do these two numbers compare with one another? If the larger number is right, how many universes with the smaller number of atoms would fit inside the larger universe?

Solution

To find the amount of the smaller number of atoms that will fit into the larger, we simply divide

Thus, we have;

10^82/10^78 = 10^4 = 10,000

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What is the slope of the line Y+2=-2(x-3)?
romanna [79]

Answer:

-2

Step-by-step explanation:

The slope is always the coefficient of x divided by the coefficient of y (if they are on opposite sides of the "=")

SO...

-2/1 = -2 = Slope

8 0
3 years ago
How do you do 5ln x= 35....?
sergejj [24]

That's going to depend a lot on what you want to "do" to it.

I'm going to assume that you want to find a number for 'x' that makes
the whole equation a true statement.  Here's how I would do it:

                                  <u> 5 ln(x) = 35</u>

Divide each side by  5:  ln(x) = 35/5
                                      ln(x)  =  7

Raise 'e' to the power of each side:    e^[ ln(x) ] = e⁷

But  e^[ ln(x) ] is just 'x'.  So x = e⁷ = <em>1,096.633...</em> (rounded)

That's the only number you can write in place of 'x'
and make the original equation true.


5 0
3 years ago
5<img src="https://tex.z-dn.net/?f=5%5E%7B56%7D%20x%205%5E%7B22%7D%20x%205%5E%7B-96%7D" id="TexFormula1" title="5^{56} x 5^{22}
Akimi4 [234]

Answer:

\frac{ {11}^{56} }{ {5}^{18} }

Step-by-step explanation:

{55}^{56}  \times  {5}^{22}  \times  {5}^{ - 96}

=  {55}^{56}  \times  {5}^{ - 74}

=  {11}^{56}  \times  {5}^{56}  \times  \frac{1}{ {5}^{74} }

=  {11}^{56}  \times  \frac{1}{ {5}^{18} }

=  \frac{ {11}^{56} }{ {5}^{18} }

8 0
3 years ago
Perpendicular distance between the base of a triangle and the opposite vertex is called _____of the triangle
Solnce55 [7]

Answer:

Median might be the answer or maybe the height?

5 0
3 years ago
Read 2 more answers
Please help im new and i need help!<br> Please help me if you onlw the answers please!!
kirza4 [7]

9514 1404 393

Answer:

  a) 2.038 seconds

  b) 5.918 meters

  c) 1.076 seconds

Step-by-step explanation:

For the purpose of answering these questions, it is convenient to put the given equation into vertex form.

  h = -4.9t² +9.2t +1.6

  = -4.9(t² -(9.2/4.9)t) +1.6

  = -4.9(t² -(9.2/4.9)t +(4.6/4.9)²) +1.6 +4.9(4.6/4.9)²

  = -4.9(t -46/49)² +290/49

__

a) To find h = 0, we solve ...

  0 = -4.9(t -46/49)² +290/49

  290/240.1 = (t -46/49)² . . . . subtract 290/49 and divide by -4.9

  √(2900/2401) +46/49 = t ≈ 2.0378 . . . . seconds

The ball takes about 2.038 seconds to fall to the ground.

__

b) The maximum height is the h value at the vertex of the function. It is the value of h when the squared term is zero:

  290/49 m ≈ 5.918 m

The maximum height of the ball is about 5.918 m.

__

c) We want to find t for h ≥ 4.5.

  h ≥ 4.5

  -4.9(t -46/49)² +290/49 ≥ 4.5

Subtracting 290/49 and dividing by -4.9, we have ...

  (t -46/49)² ≥ 695/2401

Taking the square root, and adding 46/49, we find the time interval to be ...

  -√(695/2401) +46/49 ≤ t ≤ √(695/2401) +46/49

The difference between the interval end points is the time above 4.5 meters. That difference is ...

  2√(695/2401) ≈ 1.076 . . . . seconds

The ball is at or above 4.5 meters for about 1.076 seconds.

__

I like a graphing calculator for its ability to answer these questions quickly and easily. The essentials for answering this question involve typing a couple of equations and highlighting a few points on the graph.

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<em>Additional comment</em>

I have a preference for "exact" answers where possible, so have used fractions, rather than their rounded decimal equivalents. The calculator I use deals with these fairly nicely. Unfortunately, the mess of numbers can tend to obscure the working.

"Vertex form" for a quadratic is ...

  y = a(x -h)² +k . . . .  where the vertex is (h, k) and 'a' is a vertical scale factor.

In the above, we have 'a' = -4.9, and (h, k) = (46/49, 290/49) ≈ (0.939, 5.918)

4 0
3 years ago
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