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4 years ago

PLEASE HELP WILL MARK BRAINLIEST AND 100 POINTS

Mathematics

2 answers:

Yuki888 [10]4 years ago

4 0

The Correct Answer is

Low Tide: 2:47 and 30 sec

High Tide: 9:00

A shift between Low Tide and High Tide is every 6h and 12.5

VladimirAG [237]4 years ago

3 0

Low tide will be 2:56 pm
High tide 9:00pm

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Answer with workings please

icang [17]

AB = 9 cm
BC = 6cm

CD = 7 cm
AE = 6 cm

3BC = AB
3ED = AE

  AB = AE
  BC ED
  ⁹/₃ = ⁶/ₓ
3 · 6 = 9 · x
18 = 9x
9 9
2 = x

ED = 2 cm

5 0

3 years ago

Use the Fundamental Counting Principle to find the total number of possible outcomes.

krok68 [10]

Answer:

gfhcfgjcfgjdfgjvhjvhkftfhcyjftyjvjftykftyvgjcgjfyfg gjvvgvggvgvhjvhvg

Step-by-step explanation:

vgjkcgyjvghj vhjvjvhvhjvhjvghjghjvghjvhjhvgghvghvhvhvhvhjvghjvghjvh .jvghvghj gj vjvhjvhjv .v .hvhjvhjvhvb vh vv bhjvb f

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3 years ago

Find an equation of the line passing through the pair of points. Write the equation in the form Ax +By = C.

Alexxandr [17]

9514 1404 393

Answer:

5x -y = -37

Step-by-step explanation:

One way to find the coefficients A and B is to use the differences of the x- and y-coordinates:

A = Δy = y2 -y1 = 2 -(-3) = 5

B = -Δx = -(x2 -x1) = -(-7 -(-8)) = -1

Then the constant C can be found using either point.

5x -y = 5(-7) -2 = -37

The equation of the line is ...

5x -y = -37

_____

<em>Additional comment</em>

This approach comes from the fact that the slope of a line is the same everywhere.

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\Delta y}{\Delta x}\\\\\Delta x(y-y_1)=\Delta y(x-x_1)\qquad\text{cross multiply}\\\\\Delta y(x-x_1)-\Delta x(y-y_1)=0\qquad\text{subtract y term}\\\\\Delta y(x) -\Delta x(y) = \Delta y(x_1)-\Delta x(y_1)\qquad\text{Ax+By=C form}$

The "standard form" requires that A be positive, so we chose point 1 and point 2 to make sure that was the case.

5 0

2 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago

The diagram shows the intersections of several straight roads. The avenues run parallel to each other.

Harlamova29_29 [7]

Answer:

The answer is B: 226 ft

Step-by-step explanation:

Just did the assignment on edg

5 0

3 years ago