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3 years ago

Need help 25 pts who can help me

Mathematics

2 answers:

Rashid [163]3 years ago

7 0

Answer:

Number 1 should be 2

Step-by-step explanation:

Nina [5.8K]3 years ago

6 0

Answer:

2 it guesing XD

Step-by-step explanation:

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-12x-14=-62<br> -12x = ?

KIM [24]

Answer: No solution

Step-by-step explanation: -12x-14=-62-12x

You would have to simplify the -12x but you can't so this would be a no solution.

5 0

3 years ago

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Solving polynomial(2y-4)(3y+6)

jeka57 [31]

Answer:

6y² - 24

Step-by-step explanation:

Expand. Follow FOIL method. FOIL =

First

Outside

Inside

Last.

First, multiply the first term of each parenthesis:

2y * 3y = 6y²

Next, multiply the outside terms from both parenthesis:

2y * 6 = 12y

Then, multiply the inside terms from both parenthesis:

-4 * 3y = -12y

Finally, multiply the last terms of each parenthesis:

-4 * 6 = -24

Combine like terms:

6y² + 12y - 12y - 24

6y² + (12y - 12y) - 24

6y² - 24

6y² - 24 is your answer.

3 0

4 years ago

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. Cara is playing a game. For each dart she can toss onto a board, she

Kaylis [27]

She will need 12 seconds because -20•12=-240

6 0

3 years ago

Divide by 12 nd equal 14

ICE Princess25 [194]

170 dived by 12 is 14

7 0

3 years ago

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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

3 years ago