You work for the National Weather Service and are asked to study the average daily temperatures in Anchorage, Alaska. You calculate the mean of the average daily temperatures for each month. You decide to place the information on a graph in which the date is the input and the temperature is the output. You also decide that January 1950 will correspond to the month 0 as indicated by the dot on the input scale. On a coordinate plane, quadrants 1, 2, 3, and 4 are labeled. Month is on the x-axis and degrees Fahrenheit is on the y-axis.
Determine the quadrant in which you would plot the points that correspond to the following data:
The average daily temperature for January 1936 was Negative 15 degreesF.
a.
Quadrant III
b.
Quadrant IV
c.
Quadrant II
d.
Quadrant I

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

cant open file sorry!

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

my guess would be b and d, but i didn’t actually get out the pen and paper. lol