- <u>Here</u><u>, </u><u> </u><u>we </u><u>have </u><u>to </u><u>expand </u><u>the </u><u>given </u><u>expression</u><u>s</u><u> </u><u>and </u><u>simplify </u><u>each </u><u>one </u><u>of </u><u>them</u>
<h3><u>Solution</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u></h3>
Hence, Your answer is 2mp + 4m + 3p
<h3>
<u>Solution </u><u>2</u><u> </u><u>:</u><u>-</u></h3>
Hence, your answer is 13rp + 24p + 12r
<h3><u>Solution </u><u>3</u><u> </u><u>:</u><u>-</u></h3>
Hence, your answer is 6h² + 19h + 15
<h3>
<u>Solution </u><u>4</u><u> </u><u>:</u><u>-</u></h3>
Hence, your answer is 20y² + 11y - 3
<h3>
<u>S</u><u>olution </u><u>5</u><u> </u><u>:</u><u>-</u></h3>
Hence, your answer is - 9p² - 6mp - 4nm + 6np
<span>Benford’s
law states that the probability that a number in a set has a given leading
digit, d, is
P(d) = log(d + 1) - log(d)</span>
The division property of logarithm should be use to make it
as a single logarithm
P(d) = log ( (d + 1)/ d)
So the probability that the number 1 is the leading digit
is
P(1) = log ( ( 1+1)/ 1)
P(1) = log ( 2)
<span>P(1) = 0.301</span>
Answer: A squared (written as 2 in exponent place) + B squared = C Squared
Step-by-step explanation:
This is known as Pythagorean theorem
The picture wont load for me :/
Answer:
100m
Step-by-step explanation:
just subtract them from each other
500m
- 400m
-----------
100m