2 points

What is the domain and the range

kirill [66]

Answer:

The Domain is X and Range is Y

Step-by-step explanation:

Top values are domain and bottom values are Range

6 0

3 years ago

Can anyone solve this please asap

Reika [66]

Answer:

Step-by-step explanation:

= (3 + √3) (2 - √3)

= 6 - 3√3 + 2√3 - 3

= 3 - 3√3 + 2√3

= 3 - √3 (3 - 2)

= 3 - √3

= 3 - 1.732 ...... (√3 = 1.732)

= 1.268

6 0

3 years ago

If w= -4, which inequality is true?

GenaCL600 [577]

Answer:

-5-4w>10

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help …………………………

faltersainse [42]

Answer:

u=22

Step-by-step explanation:

Triangel JKI and Triangel HKI are congruent by rule SAS (Side Angel Side)

JI and IH are congruent by rule CPCTC (Coungruent Parts of Cogruent Triangels are Cogruent).

2u=u+22

u=22

6 0

3 years ago

In an attempt to escape a desert island, a castaway builds a raft and sets out to sea. The wind shifts a great deal during the d

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

Download pdf

3 0

4 years ago