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Oxana [17]
2 years ago
5

Yellow line leaves station every 6 min and red line leaves every 8 min and they both leave at 4:15, when will they both leave at

the same time next 3 times?
Mathematics
1 answer:
zheka24 [161]2 years ago
8 0

Answer: 4:39, 5:03

Step-by-step explanation:

Given

The yellow line leaves the station every 6 min

The red line leaves the station every 8 min

They both leave at 4:15

Both trains meet again after, LCM(6,8)=24 min

i.e. at

4:15+24=4:39\\4:39+24=5:03

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Answer:

Step-by-step explanation:

Let a be the length of the altitude, then from the given triangles, applying the basic proportionality theorem,  we get

\frac{BD}{AD}=\frac{DC}{BD}

⇒(BD)^{2}=DC{\times}AD

⇒(BD)^2=9{\times}16

⇒BD=\sqrt{144}

⇒BD=12 cm

Thus, the length of altitude is: 12 cm.

Now, \frac{AB}{AD}=\frac{AC}{AB}

⇒(AB)^{2}=220

⇒AB= 14.83 cm

Also, \frac{BC}{DC}\frac{AC}{BC}

⇒(BC)^{2}=400

⇒BC=20 cm

Thus, the lengths of the legs of this triangle are 14.83 and 20 cm.

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(x - 4) to the second power

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If g(x) = x ² + 2x -8, evaluate f(-2)​
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Find dy/dx of the function y = √x sec*-1 (√x)​
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Hi there!

\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}

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Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

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f(x) = √x

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\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}

Simplify:

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