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Mathematics

1 answer:

mafiozo [28]2 years ago

3 0

Train one is going 75 m/s, train two is going 60 m/s, and train three is going 65 m/s.

Explanation:

For number one you could use the slope formula y2-y1/x2-x1, or you could simply just view it from the table that every second that’s passes the distance goes up another 75 meters. This means it would be 75 meters per second.

For the second one it’s a positive correlation that’s related meaning at every whole second (1,2,3...) the line will pass through it, so all you have to do is look at the left and see how much it’s going up by because we know it’s going up by the same amount each time.

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Please help with this

Degger [83]

All of them are wrong its 91 but if you want click or circle 90

3 0

3 years ago

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$