C3=n
N x 2 - 6 + 2
N < (6x8) - 10
Let's start by visualising this concept.
Number of grains on square:
1 2 4 8 16 ...
We can see that it starts to form a geometric sequence, with the common ratio being 2.
For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:


Thus, there are 16, 384 grains on the fifteenth square.
The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:



Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.
Answer:
KL = 27
JK = 16
MK = 30
NL = 23
m∠JKL = 132°
m∠KLJ = 22°
m∠KMJ = 54°
m∠KJL = 26°
Step-by-step explanation:
The given parameters of the quadrilateral JKLM are;
JM = 27, ML = 16, JL = 46, NK = 15, KLM = 48, JKM = 78, MJL = 22
Taking the sides as parallel, we have that quadrilateral JKLM is a parallelogram
Therefore;
KL = JM = 27
JK = ML = 16
m∠KLJ = m∠MJL = 22°
MN = NK = 15
MK = MN + NK = 15 + 15 = 30
NL = JL/2 = 46/2 = 23
m∠KJM = m∠KLM = 48°
m∠KJL = m∠KLM - m∠MJL = 48° - 22° = 26°
m∠KML = m∠JKM = 78°
m∠MKL = 180° - m∠KML - m∠KLM = 180° - 78° - 48° = 54°
m∠MKL = 54°
m∠JKL = m∠JKM + m∠MKL = 78° + 54° = 132°
m∠KMJ = m∠MKL = 54°
The first number is x axis while the second is y
Answer:
J. 19.7
Step-by-step explanation:
SOH<em>(</em><em>sine = </em><em>opposite side/hypotenuse)</em> CAH TOA
sin 80° = RS/20
=> 0.985 = RS/20
=> RS = 20 × 0.985 = 19.7