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3 years ago

What is 1/3 divided by 1/9

Mathematics

2 answers:

Anna71 [15]3 years ago

5 0

Answer:

1/27

Step-by-step explanation:

1/3 / 1/9 = 1/27

sveticcg [70]3 years ago

4 0

Answer:

Step-by-step explanation:

1/3 x (9/1) = 3

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Is .07 closer to 1/2,0 or 1

Nikitich [7]

0, because 1/2 = 0.50. therefore .07 is more closer to 0

7 0

3 years ago

The sum of two numbers is 51 and the difference is 11. what are the numbers?

stiv31 [10]

(51 - 11) : 2 = 20

20 + 11 = 31

20 and 31

6 0

3 years ago

Write down in terms of

Nata [24]

Answer:

-5

Step-by-step explanation:

Here,

t2-t1=20-25=-5

t3-t2=15-20=-5

t4-t3=10-15=-5

t5-t4=5-10=-5

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3 years ago

19 + 12k = –k − 20 solve for k

marshall27 [118]

Answer:

k=13

Step-by-step explanation:

19 + 12k = –k − 20

19+20=39=13k

k=13

8 0

3 years ago

Read 2 more answers

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