Answer:
2.................................?
Let's solve your equation step-by-step.<span><span><span><span>13</span>h</span>−<span>4<span>(<span><span><span>23</span>h</span>−3</span>)</span></span></span>=<span><span><span>23</span>h</span>−6</span></span>Step 1: Simplify both sides of the equation.<span><span><span><span>13</span>h</span>−<span>4<span>(<span><span><span>23</span>h</span>−3</span>)</span></span></span>=<span><span><span>23</span>h</span>−6</span></span>
<span>Simplify: (Show steps)</span>
<span><span><span><span><span>−7</span>3</span>h</span>+12</span>=<span><span><span>23</span>h</span>−6</span></span>Step 2: Subtract 2/3h from both sides.<span><span><span><span><span><span>−7</span>3</span>h</span>+12</span>−<span><span>23</span>h</span></span>=<span><span><span><span>23</span>h</span>−6</span>−<span><span>23</span>h</span></span></span><span><span><span>−<span>3h</span></span>+12</span>=<span>−6</span></span>Step 3: Subtract 12 from both sides.<span><span><span><span>−<span>3h</span></span>+12</span>−12</span>=<span><span>−6</span>−12</span></span><span><span>−<span>3h</span></span>=<span>−18</span></span>Step 4: Divide both sides by -3.<span><span><span>−<span>3h</span></span><span>−3</span></span>=<span><span>−18</span><span>−3</span></span></span><span>h=6</span>Answer:<span>h=6</span>
Answer:
8x-2
Step-by-step explanation:
perimeter=2l+2w
perimeter= 2(2x+3)+2(2x-4)
=4x+6+4x-8
=8x-2
The amplitude of the sine wave with RMS value of 220 V is
A = 220√2 volts.
The sine waveform is
v(t) = 220√2 sin(2πft)
where
f = 50 Hz, the frequency.
The period is
T = 1/f = 1/50 = 0.02 s
Use a graphical solution (shown below) to determine the number of times that v(t) = 220 in the interval t = [0, 0.02] s.
There are 2 instances when the voltage is 220 V in the interval t =[0, 0.02] s.
Note that 1 second is an integral multiple of 0.02 seconds.
Therefore in the interval [0,1], the number of instances when v(t) = 220 V is
(1/0.02)*2 = 100
Answer: 100