The scuba diver's position relative to sea level after the 4.6 minutes is 5.1 ft.
Given:
A scuba diver descends in the water at a rate of 23 1/2 feet per minute for 2.6 minutes.
A scuba diver ascends at a rate of 28 feet per minute for 2 minutes after seeing a big fish.
To find :
The scuba diver's position relative to sea level after the 4.6 minutes.
Solution:
Rate of a scuba diver descending in the water for 2.6 minutes =
Distance covered by a scuba diver in 2.6 minutes =
Rate of a scuba diver ascending for 2 minutes =
Distance covered by a scuba diver in 2 minutes =
The position of scuba diver after 4.6 minutes relative to sea level:
The scuba diver's position relative to sea level after the 4.6 minutes is 5.1ft.
Learn more about ascending rate and descending rate here:
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Answer:
y=2x^2+1
Step-by-step explanation:
I checked on photo maths
Answer:
See below ~
Step-by-step explanation:
Sides of a rhombus are equal.
⇒ QT = TS
⇒ x² - 4x - 10 = 6x + 14
⇒ x² - 10x - 24 = 0
⇒ x² + 2x - 12x - 24 = 0
⇒ x (x + 2) - 12 (x + 2) = 0
⇒ (x + 2)(x - 12) = 0
⇒ x = -2 or x = 12
Substitute both values and see which gives a positive value for QT :
<u>When x = -2</u> :
- QT = (-2)² - 4(-2) - 10
- QT = 4 + 8 - 10
- QT = 2
<u>When x = 12</u> :
- QT = (12)² - 4(12) - 10
- QT = 144 - 48 - 10
- QT = 86
The 2 possible answers are :
- x = -2, QT = 2
- x = 12, QT = 86
Answer:
17-6k
Step-by-step explanation:
-17k+k=6k