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3 years ago

4. If a Scooter goes 8 M/s, how fast does it go in KM/h?

Mathematics

2 answers:

Firlakuza [10]3 years ago

7 0

Answer:

28.8 km/h

Step-by-step explanation:

I used the calculator, but I don't know any formulas for changing m/s km/h.

But I hope this helps!

Agata [3.3K]3 years ago

3 0

Answer:

It would go 28.8 KM/H

Step-by-step explanation:

1 meter/second is equal to 3.6 km/h.

8 x 2.6 = 28.8

One meter is 1/1000 of a kilometer, and one second is 1/3600 of an hour. You can reference online sources for more details.

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Can someone plz help me I’m being timed hurry!!!!!!!!

diamong [38]

Answer:

1. 14 = 2 + 12

2. 16 = 4 + 12

3. 18 = 6 + 12

4. 20 = 8 + 12

Step-by-step explanation:

Super easy. All you do is replace the numbers in your table with the corresponding letter. In this case we have a table of s and f.

Example for row two: f = s + 12. Replace s with 4 ( 4 is from your s column so you would replace it with that) then solve and plug in your answer (When you solve your answer, it will go under f column).16 = 4 + 12 . f = 16, s = 4.

Formula = f = s + 12.

1. 14 = 2 + 12

2. 16 = 4 + 12

3. 18 = 6 + 12

4. 20 = 8 + 12

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3 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

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