If you're comparing 1.435 * 10^(-3) mm to 1.435 * 10^3 mm, then the more reasonable measurement is 1.435 * 10^3 mm since,
1.435 * 10^3 mm = 1435 mm = 1.435 meters
1 meter is about 3.28 feet approximately
So 1.435 meters is slightly larger (about 4.708 feet) which is a fairly reasonable distance between the tracks on a railroad.
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Something like 1.435 * 10^(-3) mm is equal to 0.001435 mm, which is far less than 1 mm.
Answer:
Step-by-step explanation:
The left hand side of the equation contains proper fractions while the right hand side of the equation contains mixed fraction. The mixed fraction can be changed to improper fraction. 1 2/3 becomes 5/3
To breakdown the left hand side of the equation, we would take lowest common factor of 5 and 15. It is 15
Considering 4/5, if 15 divides 5,the result is 3. Multiplying 3 by 4 gives 12. So it becomes
12/15
Considering 13/15, if 15 divides 15,the result is 1, Multiplying 1 by 13 gives 13. So it becomes
13/15
The equation becomes
(12 + 13)/15 = 5/3
25/15 = 5/3
Simplifying 25/15 to its lowest terms, it becomes 5/3 so
5/3 = 5/3
Answer:
Step-by-step explanation:
Given,

we know,
(f-g)(x)=f(x)-g(x)
So, here we get
(f-g)(x)
=f(x)-g(x)
=
=
=
=
So, the answer is

I believe it would be the first option.
Answer:
- B) One solution
- The solution is (2, -2)
- The graph is below.
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Explanation:
I used GeoGebra to graph the two lines. Desmos is another free tool you can use. There are other graphing calculators out there to choose from as well.
Once you have the two lines graphed, notice that they cross at (2, -2) which is where the solution is located. This point is on both lines, so it satisfies both equations simultaneously. There's only one such intersection point, so there's only one solution.
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To graph these equations by hand, plug in various x values to find corresponding y values. For instance, if you plugged in x = 0 into the first equation, then,
y = (-3/2)x+1
y = (-3/2)*0+1
y = 1
The point (0,1) is on the first line. The point (2,-2) is also on this line. Draw a straight line through the two points to finish that equation. The other equation is handled in a similar fashion.