Answer:
(a) 
(b) 
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is
and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A
The potential difference across the headlight bulbs is


(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16
-25

4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
Burning of paper is the answer