Let's solve for x.<span><span><span><span>3x</span>y</span>−<span>2x</span></span>=17
</span>Factor out variable x.<span><span>x<span>(<span><span>3y</span>−2</span>)</span></span>=17</span> Divide both sides by 3y-2.<span><span><span>x<span>(<span><span>3y</span>−2</span>)</span></span><span><span>3y</span>−2</span></span>=<span>17<span><span>3y</span>−2</span></span></span><span>x=<span>17<span><span>3y</span>−2</span></span></span>Answer:<span>x=<span>17<span><span>3y</span>−<span>2
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You applied the rule improperly. When dividing exponents, you subtract the exponent of the denominator <em>from</em> the exponent of the numerator, not the other way around. The exponent you should have gotten was 
Hi there, (6z²+z-1)(9z-5)=(6z²+z+-1)(9z+-5)=(6z²)(9z)+(6z²)(-5)+(z)(9z)+(z)(-5)+(-1)(9z)+(-1)(-5)=54z³-30z²+9z²-5z-9z+5=54z³-21z²-14z+5. Therefore, the answer is 54z³-21z²-14z+5.
I use the sin rule to find the area
A=(1/2)a*b*sin(∡ab)
1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]
we know that
A=5√3
BC=4
AB=5
then
sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°
now i use the the Law of Cosines
c2 = a2 + b2 − 2ab cos(C)
AC²=AB²+BC²-2AB*BC*cos (∡B)
AC²=5²+4²-2*(5)*(4)*cos (60)----------- > 25+16-40*(1/2)=21
AC=√21= 4.58 cms
the answer part 1) is 4.58 cms
2) we know that
a/sinA=b/sin B=c/sinC
and
∡K=α
∡M=β
ME=b
then
b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))
KE=b*sin(β)/sin(α)
A=(1/2)*(ME)*(KE)*sin(180-(α+β))
sin(180-(α+β))=sin(α+β)
A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
KE/sin(β)=KM/sin(180-(α+β))
KM=(KE/sin(β))*sin(180-(α+β))--------- > KM=(KE/sin(β))*sin(α+β)
the answers part 2) areside KE=b*sin(β)/sin(α)side KM=(KE/sin(β))*sin(α+β)Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)
