Answer:

Step-by-step explanation:
<em>Hey there!</em>
Look at the image below↓
By looking at the image we can tell the PR expression is
.
<em>Hope this helps :)</em>
Answer:
hexagonal shape
Step-by-step explanation:
i am small child from American
sorry for answer
I'm 10 year old
I am born in Nepal
but develop in America
Hi there!
v = ±
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K = 1/2mv²
Isolate for the variable "v". We can begin by dividing both sides by 1/2. (Multiply by the reciprocal, or 2):
2 · K = 2 · (1/2mv²)
2K = mv²
Continue isolating by dividing both sides by "m":
2K / m = v²
Take the square root of both sides. Remember that the solution can either be positive or negative since there are positive and negative roots.
v = ±
Answer:
a)

b) 0.09
Step-by-step explanation:
We are given the following in the question:

where B(t) gives the brightness of the star at time t, where t is measured in days.
a) rate of change of the brightness after t days.

b) rate of increase after one day.
We put t = 1

The rate of increase after 1 day is 0.09