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Misha Larkins [42]
3 years ago
11

Im having trouble finding facts that are in scientific notation. a good example is: a microbe is 1.23*10^-3 small. (this is not

a real fact I just made this up.)
Mathematics
2 answers:
BabaBlast [244]3 years ago
5 0
<span>mass of electron:9.109 times 10^-31 kg
mass of proton: 1.673 times 10-27
speed of light=3.0 times 10^8
avogadro's number 6.02 times 10^32
planck's constant: 6.63 times 10^-34</span>
I just searched on bing!
Hope it helps!
Afina-wow [57]3 years ago
4 0
Mass of electron:9.109 times 10^-31 kg
mass of proton: 1.673 times 10-27
speed of light=3.0 times 10^8
avogadro's number 6.02 times 10^32
planck's constant: 6.63 times 10^-34
just google chem constants sheet
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5x/x^2-9 + 7/x+3<br> Simplify
rjkz [21]

5x ÷ x² - 9 + 7 ÷ x + 3

Write the division as a fraction:

\frac{5x}{x^{2} }\\ - 9 + \frac{7}{x} + 3

Simplify the expression:

\frac{5}{x} - 9 + \frac{7}{x} + 3

Calculate the sum:

\frac{5}{x} - 6 + \frac{7}{x}

Write all numerators above the common denominator:

\frac{5 - 6x + 7}{x}

Add the numbers and you get the final answer:

\frac{12-6x}{x}

4 0
2 years ago
Explain: What is the solution to this system of equations? <br><br> 2x+y=20 <br> 6x-5y=12
bazaltina [42]

Answer:

solution set is (x,y) = (7,6)

Step-by-step explanation:

solving by substitution method

2x +y=20--------------1

6x-5y=12---------------2

from equation 1, solve for y

2x+y=20

y= 20-2x------equation 3

adding value of y in equation 2

6x-5y=12

6x-5(20-2x)=12

6x-100+10x=12

16x= 12+100

16x= 112

x= 112/16

x=7

adding value of x in equation 3

y= 20-2x

y= 20- 2(7)

y=20-14

y=6

so solution set (x,y) = (7,6)

3 0
3 years ago
Miki has 104 nickels and 88 dimes. She wants to divide her coins into groups where each group has the same number of nickels and
pochemuha

Given :

Miki has 104 nickels and 88 dimes.

She wants to divide her coins into groups where each group has the same number of nickels and the same number of dimes.

To Find :

Largest number of groups she can have .

Solution :

In the given question we need to find the largest number of groups she can have i.e we have to find the LCM of 104 and 88 .

Now , factorizing both of them , we get :

104=1\times 2\times 2\times 2\times 13

88=1\times 2\times 2\times 2\times 11

Form above , we can say that common factors are :

HCF=2\times 2\times 2=8

Therefore , the largest number of groups she can have is 8 .

Hence , this is the required solution .

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Kazeer [188]
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