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m_a_m_a [10]
3 years ago
11

Two passenger trains 190 miles apart are headed toward each other on the same track—one at 50 mph and the other at 80 mph. Ironm

an responds to the danger by flying back and forth between the trains, hoping to stop them before disaster strikes. Unfortunately, when he reaches each train, an invisible force repulses him back, jolting him in the opposite direction (toward the other train).  If Ironman maintains a constant speed of 150 mph the entire time—while flying back and forth between the trains—how far will he have flown when the trains collide?
Mathematics
1 answer:
frozen [14]3 years ago
5 0
190-(50+80)t=0

190-130t=0

130t=190

t=19/13  so the trains will remain apart for 19/13 hours.

Since ironman is flying at a constant rate of 150mph he will fly:

150(19/13)≈219.23 miles in that time period while the trains are apart.
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3 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

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\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

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Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

6 0
3 years ago
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