Question

Let f(x)=100−10+e−0.1x .

What is f(−2) ?


Enter your answer, rounded to the nearest tenth, in the box.
please i need this answer really quick

Answer 1

What are you solving for?
If you are solving for × then I hope this helps

Let's solve for x.

f(x)= 100−10+e−0.1x

Step 1: Add 0.1x to both sides.

xf+0.1x=−0.1x+e+90+0.1x

xf+0.1x = e+90

Step 2: Factor out variable x.

x(f+0.1) = e+90

Step 3: Divide both sides by f+0.1.

x(f+0.1)/ f+0.1 =e+90/ f+0.1

x=e+90/ f+0.1

Answer:

x= e+90/ f+0.1

Answer 2

f(-2) = 87.8

Explanation:

Given : function
`f(x)=100-10e^((-0.1x))`

We have to find f(-2)

Consider the given function
`f(x)=100-10e^((-0.1x))`

We have to find f (-2) that is value of function f(x) at x = -2

Put x = -2 in function given , we have,

`f(-2)=100-10e^(-0.1\left(-2\right))`

`\mathrm{Apply\:rule}\:-\left(-a\right)=a`

`=100-10e^(0.1\cdot \:2)`

Multiply , we have,

`=100-10e^(0.2)`

We know
`e^(0.2)=1.22140\dots`

So,
`10e^(0.2)=12.21403\dots`

Subsitute and solve, we have,

`f(-2)=100-12.21403\dots=87.78597\dots`

Thus, f(-2) = 87.8