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✔cos^2A+sin^2A=1
✔1-cos^2A=sin^2A
✔cos2A=cos^2A-sin^2A
✔sin2A=2.sinA.cosA
secA=1/cosA
tgA=sinA/cosA
sin^2A/1/cos^2A-sin^2A/cos^2A
sin^2A.cos^2A/cos2A
2.sin^2A.cos^2A/cos2A
sin2A.2.sin2A/cos2A
tg2A.2.sin2A
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°
-9 is the slope of the line.
The amount of metal used to make the can = surface area of cylindrical can, which is: D. 42π in².
<h3>What is the Surface Area of a Cylinder?</h3>
A can is a cylinder.
Surface area of a cylinder = 2πrh + 2πr²
Thus:
Amount of metal used to make the can = surface area of the can
r = 6/2 = 3 in.
h = 4 in.
Therefore:
Surface area = 2×π×3×4 + 2×π×3²
Surface area = 24π + 18π
Surface area = 42π in².
Learn more about the surface area of a cylinder on:
brainly.com/question/14657844
