Answer:
≈ 1833
Step-by-step explanation:
To find ratio of proton mass to electron mass, we have to divide.
The numbers are given in <em>scientific notation</em>.
Let a number be 
and another be 
, when we divide, we will follow the rule shown below:
Now, we use the information given to find the ratio:
This means we can find the number by taking 4 decimal places to the right, so that would becomes:
The approximate ratio is 1833 [mass of proton is around 1833 times heavier than mass of electron]
Answer:
1.256, 1.265, and 1.268
Step-by-step explanation:
Your answer would be option B. 2y² - y - 6 = 0. This is because if you were to substitute x = y² - 1 into the equation 2x - y = 4, you would get 2(y² - 1) - y = 4, which expands into 2y² - 2 - y = 4, and then simplifies to 2y² - y - 6 = 0.
I hope this helps!
Answer:
The center of the circle is 
.
Step-by-step explanation:
The center of the circle is the midpoint of the segment between the endpoints. We can determine the location of the center by this vectorial expression:
(1)
Where:
- Center.
, 
- Location of the endpoints.
If we know that 
and 
, then the location of the center of the circle is:
The center of the circle is .
Answer: line CD (fourth choice)
====================================================
Let's go through the choices one by one
1) Segment AB is a radius, which is not a secant line. We can rule this out.
2) line DE is a tangent line which only touches the circle at exactly one spot. We need something that cuts the circle in 2 spots for it to be a secant line. This can also be crossed off the list.
3) segment HG is a chord, which is fairly close to a secant line, but it must extend infinitely in both directions. In other words, it needs to be a line instead of a line segment. This is crossed off the list.
4) line CD is a secant line. It is a geometric line in that it goes on forever in both directions (it's not a segment, and not a ray either). Also, this line crosses the circle at 2 points, which contrasts it from a tangent line. This is the answer we want.
