Answer:
C) Rotated parallel lines always remain parallel lines.
Step-by-step explanation:Rotated parallel lines remain parallel lines.
Rotation is a rigid transformation, so the lines remain parallel.
I can't give the exact answer as the base pay and weekly hours arn't stated but here's how to solve letters in parentheses are place holders just plug in the numbers accordingly :)
Step-by-step explanation:
<em>A</em><em>.</em>
figures needed, base pay (p) total hours worked (t) half
Saturday
p÷2= the ½ in time and a half (x)
p+x= 1 hr Saturday pay (s)
s+2= r (to cover the 30m)
7:00am to 11:30am 4hr 30m
12:00pm to 4:00pm 4hr
add them
8hr 30m
(s×8)+r= how much you were paid on Saturday
Sunday
p×2= h dubble pay
Calculate how many hrs worked on Sunday and apply the same logic as for Saturday appropriately divide the time if it doesn't come out as an even hour.
Final
Add Saturday and Sunday pay together to get weekend pay.
<em>B</em><em>.</em>
Same formula hours worked times hourly pay.
Add the weekday and weekend pay together and thats pt B.
Hope this helps!
Answer:
1. Domain: -2, 0, 4; Range: -7, -1, 0, 3
2. Domain: -4, -1, 2, 3 Range: -5, 1, 8
Step-by-step explanation:
Domain and range are the just the x and y value in function table or ordered pair.
Lets imagine the shape
M
/\
/ | \
/ | \
/ | \
P /___ |___\ N
O
Now in If we take â†MOP and â†MON
As MO ⊥ NP so â MON=â MOP
NO=NP (given)
And MO is a common side
so by side angle side rule of congruency
â†MOP and â†MON are congruent
so MP is congruent to MN
Answer:
A = 0.75 gram or 1 gram
Step-by-step explanation:
The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole number).
We can use the following formula for half-life of 
to find out how much is left from the original sample after 2,292 years:
where:
<em>A</em> is the amount left of an original gram sample after <em>t</em> years, and
is the amount present at time <em>t</em> = 0.
The half-life of is the time <em>t</em> at which the amount present is one-half the amount at time <em>t </em>= 0.
If 1 gram of is present in a sample,
Solve for A when t = 2,292:
Substituting = 1 gram into the decay equation, and we have:
A = 0.75 g or 1 g
