The table shows the dates, times, and high and low tide heights. Cresap Bay Tide Table
Date
Time
Low/high tide
Thursday, Feb. 7
1:26 a. m.
2.6 ft (low)
7:26 a. m.
6.4 ft (high)
2:29 p. m.
–0.8 ft (low)
9:21 p. m.
4.8 ft (high)
Friday, Feb. 8
2:24 a. m.
1.5 ft (low)
8:22 a. m.
6.5 ft (high)
3:16 p. m.
–1.6 ft (low)
10:02 p. m.
5.1 ft (high)

On what date and time was the lowest tide?
Thursday, Feb. 7, at 2:29 p. m.
Friday, Feb. 7, at 9:21 p. m.
Friday, Feb. 8, at 2:24 a. m.
Friday, Feb. 8, at 3:16 p. m.

i dony know

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles.

i took the test

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

i need points

step-by-step explanation:

sorry bubs