Mr. mole left his burrow that lies 7 meters below the ground and started digging his way deeper into the ground, descending at a constant rate. after 6 minutes, he was 16 meters below the ground. let  a(t)a(t)a, left parenthesis, t, right parenthesis  denote mr. mole's altitude relative to the ground  aaa  (measured in meters) as a function of time  ttt  (measured in minutes).what's the functions formula?

A(t)=−1.8t−4.5

Step-by-step explanation:

Mr. Mole is descending at a constant rate, so we are dealing with a linear relationship.

We know that Mr. Mole descended at a rate of 1.81.81, point, 8 meters per minute, so the slope \green mmstart color green, m, end color green is \green{-1.8}−1.8start color green, minus, 1, point, 8, end color green, and our function looks like A(t)=\green{-1.8}t+\pink bA(t)=−1.8t+bA, left parenthesis, t, right parenthesis, equals, start color green, minus, 1, point, 8, end color green, t, plus, start color pink, b, end color pink.

We also know that after 555 minutes, Mr. Mole was 13.513.513, point, 5 meters below the ground, which means that A(5)=-13.5A(5)=−13.5A, left parenthesis, 5, right parenthesis, equals, minus, 13, point, 5. We can substitute this into the formula of the function to find \pink bbstart color pink, b, end color pink:

\qquad\begin{aligned}A(5)&=-13.5\\ \green{-1.8}\cdot5+\pink b&=-13.5\\ -9+\pink b&=-13.5\\ \pink b&=\pink{-4.5}\end{aligned}  

A(5)

−1.8⋅5+b

−9+b

b

​    

=−13.5

=−13.5

=−13.5

=−4.5

​  

This means that Mr. Mole's initial altitude is 4.54.54, point, 5 meters below the ground.

AKA the answer is A(t)=−1.8t−4.5

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: