M=(y2-y1)/(x2-x1)
m=(-4-4)/(3-1)
m=-8/2
m=-4
so the answer should be first one
You have the correct approximate value of k. The part after that is where things get a bit strange.
You don't plug in t = 5 again. We don't know the value of t yet. We're trying to solve for it.
We do know u(t) is 135 as we want the temperature to be 135 degrees. Plug this in and isolate t.
So
u(t) = 70+380e^(-0.1004183888t)
135 = 70+380e^(-0.1004183888t)
135-70 = 380e^(-0.1004183888t)
65 = 380e^(-0.1004183888t)
380e^(-0.1004183888t) = 65
e^(-0.1004183888t) = 65/380
e^(-0.1004183888t) = 13/76
-0.1004183888t = ln(13/76)
-0.1004183888t = -1.7657839828248
t = -1.7657839828248/(-0.1004183888)
t = 17.5842692152893
So it takes about 17.58 minutes for the item to cool to the desired temp
There are 4 notebooks left. Hope this helps :)
Answer:
The measure of angle KLM is 62°
Step-by-step explanation:
* Lets revise the properties of the rhombus
- The rhombus has 4 equal sides in length
- Every two opposite angles are equal in measure
- The two diagonals bisect each other
- The two diagonals perpendicular to each other
- The two diagonals bisect the vertices angles
* Lets solve the problem
∵ JKLM is a rhombus
∴ m∠MJK = m∠KLM ⇒ opposite angles in the rhombus
∵ JL and KM are diagonals in the rhombus and intersect each
other at N
∴ JL bisects ∠MJK
∴ m∠KJL = m∠MJN
∵ m∠KJL = (2x + 5)°
∵ m∠MJN = (3x - 8)°
∴ 2x + 5 = 3x - 8 ⇒ subtract 2x from both sides
∴ 5 = x - 8 ⇒ add 8 to both sides
∴ 13 = x
∴ The value of x = 13
∵ m∠KJL = (2x + 5)° ⇒ substitute the value of x
∴ m∠KJL = 2(13) + 5 = 26 + 5 = 31°
∵ m∠KJL = 1/2 m∠MJK
∴ m∠MJK = 2 m∠KJL
∴ m∠ MJK = 2 × 31° = 62°
∵ m∠MJK = m∠KLM ⇒ opposite angles in the rhombus
∴ m∠KLM = 62°
