Answer:
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<em>m∠LNM = 54°</em>
<u><em>Here is why:</em></u>
In this photo there are two important angles with very important features, a central angle and an inscribed angle.
A central angle is an angle that is in the center of the circle, so angle P is a central angle. The arc that is associated with this angle is going to be the same measure as the central angle. I have labeled this in the photo below as <u>blue</u>.
An inscribed angle is an angle that lies on the circle, so angle N is an inscribed angle. The arc that is associated with this angle will be double the amount of the inscribed angle, or the angle is half of the measure of the arc. I have labeled this in the photo below as <u>red</u>.
Since we know that the central angle is 108°, with what we know about central angles we know that arc LM is going to be 108° as well.
We also know that an inscribed angle is half the amount of the arc so...
108 ÷ 2 = 54
<em><u>m∠LNM = 54°</u></em>
Answer:
a)z1 +z2 =z2 + z1 ...proved.
b) z1 + ( z2+ z3 )=(z1+z2)+z3 ... proved.
Step-by-step explanation:
It is given that there are three vectors z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3. Now, we have to prove (a) z1 + z2 = z2 + z1 and (b) z1 + (z2 +z3) = (z1 + z2) + z3.
(a) z1 + z2 = (a1 +ib1) + (a2+ ib2) = (a1 +a2) + i(b1 +b2) {Adding the real and imaginary parts separately}
Again, z2 + z1 =(a2 +ib2) + (a1 +ib1) = (a2 +a1) + i(b2 +b1) {Adding the real and imaginary parts separately}
Hence, z1 +z2 =z2 + z1 {Since, (a1 +a2) = (a2 +a1) and (b1 +b2) = (b2 +b1)}
(b) z1 + ( z2+ z3 ) = [a1 + ib1] + [(a2 + a3 ) + i(b2 + b3 )] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}
Again, (z1+z2)+z3 = [(a1+a2) +i(b1+b2)]+[a3+ib3] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}
Hence, z1 + ( z2+ z3 )=(z1+z2)+z3 proved.
Answer:
9 - 5x = 7
Step-by-step explanation:
Nine (9) less (-) then the product (*) of 5*x, equals (=) to 7
9 - 5x = 7
