You have one type of chocolate that sells for $6.00/lb and another type of chocolate that sells for $9.70/lb. You would like to have 11.1 lbs of a chocolate mixture that sells for $7.50/lb. How much of each chocolate will you need to obtain the desired mixture? You will need lbs of the cheaper chocolate and lbs of the expensive chocolate.

lbs of the cheaper chocolate => x = 6.6 pounds(Ibs)

lbs of the expensive chocolate => y = 4.5 pounds(Ibs)

Step-by-step explanation:

Let's the :

lbs of the cheaper chocolate => x

lbs of the expensive chocolate => y

You would like to have 11.1 lbs

Hence:

x + y => 11.1

x = 11.1 - y

You have one type of chocolate that sells for $6.00/lb and another type of chocolate that sells for $9.70/lb. of a chocolate mixture that sells for $7.50/lb.

Hence:

$6.00 × x + $9.70 × y = $7.50 × 11.1

6x + 9.7y = 83.25

Hence, we substitute 11.1 - y for x

6(11.1 - y) + 9.7y = 83.25

66.6 -6y + 9.7y = 83.25

-6y + 9.7y = 83.25 - 66.6

= 3.7y => 16.65

y = 16.65/3.7

y = 4.5 pounds

x = 11.1 - y

x = 11.1 - 4.5

x = 6.6 pounds

Hence:

You will need 4.5 lbs of the cheaper chocolate and 6.6 lbs of

the answer is 13 cm

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.