26+0.08x=8+0.13x
26-8=0.13x-0.08x
18=0.05x
x=18/0.05=360 min
6x = 15
I apologize if this is wrong!
Answer:
hypotenuse
Step-by-step explanation:
the other sides are known as legs
For the answer to the question above,
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3
<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>
<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>
<span>1 + 9x + 36x^2 + 84x^3 </span>
<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>
<span>(1+x) = 0.98 </span>
<span>x = 0.98 - 1 = -0.02 </span>
<span>Substituting: </span>
<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>
<span>approximation = 0.834 </span>
<span>Checking the real value in your calculator: </span>
<span>(0.98)^9 = 0.834 </span>
<span>So you have approximated correctly. </span>
<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>
<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>
<span> (0.98)^9 = 0.8337477621 </span>
<span>So it is correct to 4</span>
1) 6x-7y=-11
2) x=-2
3) y=-4x+1
4) y-5=3(x-1)
