Answer: B: n^2+6n+1
Step-by-step explanation:
A=n
B=2n+6
C=n^2-1
AB-C
n(2n+6)-n^2-1
2n^2+6n-n^2+1
n^2+6n+1
In 1944 Elion joined the Burroughs Wellcome Laboratories (now part of GlaxoSmithKline (a company that makes prescription medicines)). There she was first the assistant and then the colleague of Hitchings, with whom she worked for the next four decades. Elion and Hitchings developed an array (variety) of new drugs that were effective against leukemia, autoimmune disorders, urinary-tract infections, gout, malaria, and viral herpes. Their success was due primarily to their innovative (characterized by new or unique) research methods. Rather than using the trial-and-error approach used by previous pharmacologists, Elion and Hitchings examined the difference between the biochemistry of normal human cells and that of cancer cells, bacteria, viruses, and other pathogens (disease-causing agents). They used this information to create drugs that could target a particular pathogen without harming the human host's normal cells. Their methods enabled them to eliminate much of the guesswork and wasted effort typical in previous drug research.
Answer:
speed of Jada = 6km per hour which is not equal to
speed of Andre = 8km per hour
They did not run at same speed
Step-by-step explanation:
formula of speed = distance covered/ time taken
For Andre
Distance = 2 KM
time = 15 mins
60 mins = 1 hour
15 mins = 1/60 * 15 hour = 0.25 hours
thus,
speed = 2km/0.25 hour = 8 km per hour
Similarly for Jada
Distance = 3 KM
time = 20 mins
60 mins = 1 hour
20 mins = 1/60 * 20 hour = 1/3 hours
thus,
speed = 2km/(1/3)hour = 6 km per hour
Now we have
speed of Jada = 6km per hour which is not equal to
speed of Andre = 8km per hour
Answer:
25 posts
Step-by-step explanation:
So the number of fence post would be the total length of the log divided by the length of each post. As the log is 16m and is corrected to the nearest metre, it could possibly be 16.499m. As for the post that is 70 cm long and corrected to the nearest 10cm, it may as well be 65 cm (or 0.65m) each post
So the max number of fence point once can possibly cut from the log would be
16.499 / 0.65 = 25 posts
