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3 years ago

How do you do this? It's due tomorrow

Mathematics

2 answers:

Korvikt [17]3 years ago

4 0

Lets think they have cards, Alex = x , Gil = y and Meg = z. Then,
x+y+z = 740

Alex had 20 fewer cards than Gil. So,
x+20 = y

Alex had 120 fewer cards than Meg. So,
x+120 = z

x+(x+20)+(x+120) = 740
3x + 140 = 740
x= 200

y= 220
z= 320

<span>So Alex had 200 cards, Gil had 220 cards and Meg had 340 cards</span>

belka [17]3 years ago

3 0

A=alex
g=gil
m=meg

they share 740 cards
a+g+m=740

alex has 20 fewer than gil and 120 fewer than meg (women rights?)
a=g-20 and a=m-120

so

a=g-20 and a=m-120
add 20 to bot sides for the left one and 120 to boht isdes for the right one
a+20=g and a+120=m

subsitute a+20=g for g and a+120=m for m

a+g+m=740
a+a+20+a+120=740
3a+140=740
minus 140 both sides
3a=600
divide both sies by 3
a=200

alex has 200 cards

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First we define two generic vectors in our $\mathbb{R}^2$ space:

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Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

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As second condition we have that:

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$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

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