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3 years ago

Write an expression for the operation described below. the quotient of z and y

Mathematics

1 answer:

Vsevolod [243]3 years ago

7 0

Answer:

z/y

Step-by-step explanation:

the quotient of z and y

this does no require any work just understand the question

the quotient of z and y means z divide by y

z/y

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nikdorinn [45]

Answer:

R = 7,34 km

Ф = 63,43° north of east

Step-by-step explanation:

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

Please see images for each displacement in attached file.

Displacement A: 2.50 km and 45° north of west

Ax= AcosФ = 2.50cos45° = 1.77 km (neg. direction)

Ay= AsenФ = 2.50sen45° = 1.77 km (pos. direction)

Displacement B: 4.70 km and 60° south of east

Bx= BcosФ = 4.70cos60° = 2,35 km (pos. direction)

By= BsenФ = 4.70sen60° = 4,07 km (neg. direction)

Displacement C: 1.30 km and 25° south of west

Cx= CcosФ = 1.30cos25° = 1,18 km (neg. direction)

Cy= CsenФ = 1.30sen25° = 0.55 km (neg. direction)

Displacement D: 5.10 km due east (0°)

Dx= DcosФ = 5.10 km (pos. direction)

Dy= DsenФ = 0 km

Displacement E: 1.70 km and 5° east of north

Ex= EsenФ = 1.70sen5° = 0,15 km (pos. direction)

Ey= EcosФ = 1.70cos5° = 1.69 km (pos. direction)

Displacement F: 7.20 km and 55° south of west

Fx= FcosФ = 7.20cos55° = 4.13 km (neg. direction)

Fy= FsenФ = 7.20sen55° = 5.90 km (neg. direction)

Displacement G: 2.80 km and 10° north of east

Gx= GcosФ = 2.80cos10° = 2.76 km (pos. direction)

Gy= GsenФ = 2.80sen10° = 0.49 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx + Ex + Fx + Gx

Ry = Ay + By + Cy + Dy + Ey + Fy + Gy

Therefore,

Rx= (-1.77)+(2.35)+(-1.18)+(5.10)+(0.15)+(-4.13)+(2.76)

Ry= (1.77)+(-4.07)+(-0.55)+(0)+(1.69)+(-5.90)+(0.49)

Note: minus (-) symbol to negative directions

Rx = 3.28 km

Ry = - 6.57 km

Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

$R = \sqrt{Rx^{2} +Ry^{2} }$

$R =\sqrt{(3.28)^{2}+(-6.57)^{2} }$

R = 7,34 km

To find the direction of the resultant:

tanФ = $\frac{Ry}{Rx}$

tanФ = $\frac{6.57}{3.28}$

tanФ = 2.00

Ф = tan-1 (2.00)

Ф = 63,43° north of east

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