Our math team had m students last year, but this year it has already n students! By what percent did the number of students grow
1 answer:
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Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>Step-by-step explanation:
Given the function j(x) = 11.6 and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6 and k(x) =
are inverse functions.