We know that
area of lateral faces of the <span>right triangular prism=perimeter of base*height
perimeter of the base=[3+4+5]----> 12 cm
height of the prism=5 cm
so
</span>area of lateral faces of the right triangular prism=12 cm*5 cm---> 60 cm²
the answer is
60 cm²
Hey there! :D
First, find the slope.
m= (y2-y1)/(x2-x1) I'm using points (0,3) and (1,5)
m=(5-3)/(1-0)
m= 2/1
m=2
Now, plug that into point slope. Use point (1,5) m=2
y-y1= m(x-x1)
y-5= 2(x-1)
Distribute.
y-5=2x-2
Add 5 to both sides.
y=2x+3 == equation of the line in slope intercept form. y=mx+b
I hope this helps!
~kaikers
By Direct Proof :
<span><span>1.(∼H(x)∨∼S(x))→(P(x)∨L(x))</span><span>1.(∼H(x)∨∼S(x))→(P(x)∨L(x))</span></span>
<span><span>2.P(x)→E(x)</span><span>2.P(x)→E(x)</span></span>
<span><span>3.∼E(x)</span><span>3.∼E(x)</span></span>
<span><span>−−−−−−−−−−−−−−−−−−−−−</span><span>−−−−−−−−−−−−−−−−−−−−−</span></span>
<span><span>4.H(x)</span><span>4.H(x)</span></span>
<span><span>5.P(x)→E(x)≡∼P(x)∨E(x)</span><span>5.P(x)→E(x)≡∼P(x)∨E(x)</span></span> by Material Implication
<span><span>6.∼P(x)</span><span>6.∼P(x)</span></span> , #5 and #3 by Disjunctive Syllogism
<span><span>7.∼P(x)∨∼L(x)</span><span>7.∼P(x)∨∼L(x)</span></span> , #6 by Addition ( I just add <span>∼∼</span>L(x))
Since #7 is logically equivalent to <span><span>∼(P(x)∨L(x))</span><span>∼(P(x)∨L(x))</span></span> by De Morgan's Law,
<span><span>8.∼(∼H(x)∨∼S(x))</span><span>8.∼(∼H(x)∨∼S(x))</span></span> , #1 and #7 by Modus Tollens.
Distributing the <span>∼∼</span>, we'll have,
<span><span>9.H(x)∧S(x)</span><span>9.H(x)∧S(x)</span></span> by De Morgan's and Double Negation
<span><span>10.H(x)</span><span>10.H(x)</span></span> by Simplification <span>■</span>
