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# Calculate the approximate value of $1.01$$1.05$^2 e^1.1+$1.01$ln$1.05$ using the verification of the three-variable function

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## Answer to a math question Calculate the approximate value of $1.01$$1.05$^2 e^1.1+$1.01$ln$1.05$ using the verification of the three-variable function

Andrea
4.5
Let's denote the three variables as follows:

x = 1.01 , y = 1.05 , z = 1.1

The given expression can be rewritten as:

f$x, y, z$ = x y^2 e^z + x \ln$y$

Substitute the given values of x , y , and z into the function f$x, y, z$ :

f$1.01, 1.05, 1.1$ = 1.01 \times 1.05^2 \times e^{1.1} + 1.01 \times \ln$1.05$

Now, calculate the values:

f$1.01, 1.05, 1.1$ = 1.01 \times 1.1025 \times 3.004166 + 1.01 \times \ln$1.05$

f$1.01, 1.05, 1.1$ = 1.1426 + 1.01 \times \ln$1.05$

Now, use the third variable function to simplify:

Let's define another function as g$t$ = 1.01 \times \ln$t$

So, g$1.05$ = 1.01 \times \ln$1.05$

Therefore, the final function becomes:

f$1.01, 1.05, 1.1$ = 1.1426 + g$1.05$

Now calculate the value of g$1.05$ :

g$1.05$ = 1.01 \times \ln$1.05$ \approx 1.01 \times 0.04879 \approx 0.0492769

Finally, substitute this value into the original function:

f$1.01, 1.05, 1.1$ = 1.1426 + 0.0492769 \approx \boxed{1.19188}

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