Please help it’s urgent . Grade 11 homework

Mathematics

1 answer:

AlexFokin [52]2 years ago

5 0

Answer:

(l) y=2x-2

(ll) y=4x+1

(lll) y=5x+13

(lV) y=-3x-6

(V) y=-5x-17

(Vl) y=(2/3)x-3

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A father and his son walked 240 meters. During this trip, the father made 100 fewer steps than did his son. Find the distance ea

Alexxandr [17]

<h2>Hello!</h2>

The answer is:

The father's step is 80 cm

The son's step is 60 cm.

To solve the problem, we need to write equations in order to create a relation between the father's and son's steps, and the distance covered by them.

We know thay they covered 240 meters, and the father's step is 20 cm (0.2m)greater than his son's step, so, we can write the following relation:

$NumberOfSteps_{Son}=\frac{240m}{x}$

and,

$NumberOfSteps_{Father}=\frac{240m}{x+0.2m}$

Now, if the know that the father made 100 fewer steps than his son, we have:

$NumberOfSteps_{Son}-100=\frac{240m}{x+0.2m}$

Then, substituting the first equation into the second equation, we have:

$\frac{240m}{x}-\frac{240m}{x+0.2m}=100\\\\\frac{240m(x+0.2m)-240mx}{x(x+0.2m)}=100\\\\240mx+0.2m*240m-240mx=(x)(x+0.2m)*100\\\\48m^{2}=100*(x^{2}+0.2mx)=100x^{2} +20mx\\\\48m^{2}=100x^{2}+20mx\\\\100x^{2}+20mx-48m^{2}$

We have a quadratic equation:

$100x^{2}+20mx-48m^{2}=0$

Where,

$a=100\\b=20\\c=-48$

So, solving it applying the quadratic formula, we have:

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

$\frac{-20+-\sqrt{20^{2} -4*100*-48} }{2*100}=\frac{-20+-\sqrt{400+19200} }{200}\\\\\frac{-20+-\sqrt{400+19200} }{200}=\frac{-20+-\sqrt{19600} }{200}\\\\\frac{-20+-\sqrt{19600} }{200}=\frac{-20+-(140)}{200}\\\\x_{1}=\frac{-20+140}{200}=0.6\\\\x_{2}=\frac{-20-140}{200}=-0.8$