Jackson invested $590 in an account paying an interest rate of 6 % compounded

Someone help, please and thank you

taurus [48]

Answer:

6√2

Step-by-step explanation:

Given,

θ = 45

Opposite side = 6

To find : - Hypotenuse

Formula : -

sin θ = Opposite side / Hypotenuse

[ The value of sin 45 = 1 / √2 ]

sin 45 = 6 / Hypotenuse

1 / √2 = 6 / Hypotenuse

Cross multiply,

Hypotenuse = 6√2

8 0

2 years ago

Read 2 more answers

In triangle XYZ, let angle Y be a right angle, mZ = 54°, and YZ = 361. Find the area of the

laiz [17]

Answer:

89,709

Step-by-step explanation:

∡Z= 54°

tan 54 = XY / 361

XY = 361 · tan 54°

XY = 497

A = 1/2 (361) (497)

5 0

3 years ago

The scatter plot shows the sizes and annual rents of some office spaces in the downtown area of a city. What would a line of bes

strojnjashka [21]

Give me a second please. Will have done in a second

8 0

3 years ago

What is the general term equation, an, for the arithmetic sequence 9, 5, 1, −3, . . . , and what is the 21st term of this sequen

marin [14]

Hello :
<span>the general term equation, an, for the arithmetic sequence is :
An = A1+d(n-1)
A1 : the first term and : d the common diference
in this exercice :
A1 = 9
d =(-3)-(1) =(1)-(5) =(5) -(9) = -4
An = 9-4(n-1)
An = -4n +13</span>

4 0

3 years ago

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

7 0

2 years ago