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2 years ago

I need someone to go out with me i am 18

Mathematics

2 answers:

Anna [14]2 years ago

7 0

Answer:

Step-by-step explanation:

you're really on brainly trying to find a s/o?

KatRina [158]2 years ago

5 0

Answer:

get tinder....

Step-by-step explanation:

You might be interested in

Determine a series of transformations that would map Figure RR onto Figure SS.

nadezda [96]

Answer:

transformation

Step-by-step explanati

3 0

2 years ago

If 5x-26=x+50, then the value of x is

horsena [70]

Let's work this problem out!

First: 5(x) - 26 - x + 50 = ?

Second: 4(x) - 76 = ?

Third: 4 = ?
↑
(Solve Step 3)

Fourth: x - 19 = 0
↑
(Add the number 19 to each of your sides for Step 4)

Finally we get to your answer

19 + 0 = 19 ; 19 - 19 = 0

So, therefore the value of "x" is 19

$x = 19$

3 0

2 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

Please help with this!!!

RSB [31]

Answer:

Step-by-step explanation:

So to graph an inverse of the function given we have to switch the x and y values of the function:

So Ill do some points and the rest of the points the same process is applicable.

So our first point is (-1,1)

So our inverse point would be (1,-1)

The second point is (-2,2)

So our inverse point would be (2,-2)

The third point is (-3,3)

So our inverse point would be (3,-3)

Hope this helps!

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=1%20%2B%205%20%5Ctimes%206y" id="TexFormula1" title="1 + 5 \times 6y" alt="1 + 5 \times 6y" al

Levart [38]

Answer:

30y+1

Step-by-step explanation:

1+5x6y

1+30y

30y+1

6 0

2 years ago