stella94 stella94
  • 03-04-2019
  • Mathematics
contestada

please help with this math question​

please help with this math question class=

Respuesta :

Аноним Аноним
  • 03-04-2019

For the first question, you just need to plug the values and solve for k:

[tex]11 = \dfrac{3000}{500}+k \iff 11=6+k \iff k=11-6=5[/tex]

Now we know that [tex]k=5[/tex], so the cost of 200 cards will be

[tex]y = \dfrac{3000}{200}+5 = 15+5=20[/tex]

Finally, if we want the cost to be 7 we're asking

[tex]7=\dfrac{3000}{x}+5 \iff \dfrac{3000}{x}=2 \iff x = \dfrac{3000}{2}=1500[/tex]

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