Find an equation for the tangent to the curve at the given point.
7) $f(x)=10\sqrt{x}-x+7,(100,7)$

Step 1 ：Given the function $f(x)=10\sqrt{x}-x+7$ and the point $(100,7)$.

Step 2 ：First, we need to find the derivative of the function $f(x)$.

Step 3 ：The derivative of $f(x)$ is $f^{\prime }(x)=-1+\frac{5}{\sqrt{x}}$.

Step 4 ：Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. So, $m=f^{\prime }(100)=-1+\frac{5}{\sqrt{100}}=-\frac{1}{2}$.

Step 5 ：Then, we use the point-slope form of the line equation to find the equation of the tangent line. The point-slope form is $y-y_1=m(x-x_1)$.

Step 6 ：Substitute the given point $(100,7)$ and the slope $m=-\frac{1}{2}$ into the point-slope form, we get $y-7=-\frac{1}{2}(x-100)$.

Step 7 ：Solve for y to get the equation of the tangent line: $y=57-\frac{x}{2}$.

Step 8 ：Final Answer: The equation of the tangent line to the curve $f(x)=10\sqrt{x}-x+7$ at the point $(100,7)$ is $\boxed{{\displaystyle y=57-\frac{x}{2}}}$.