0.6 −0.2 =1.5(a+c(b+0.8)) =−2.5(b−0.4(1.2−1.5a)) consider the system of equations above, where ccc is a constant. for which value of ccc are there no (a, b)(a, b)left parenthesis, a, comma, b, right parenthesis solutions?

For c equal to 5/3

Step-by-step explanation:

For solve these, we are going to find and expression for a terms of c and determine for which value of c it is not possible to find a value for a. So, we have 2 equations:

0.6 = 1.5 ( a + c ( b + 0.8 ) )    eq. 1

-0.2 = -2.5 (b - 0.4 ( 1.2 - 1.5a) )    eq. 2

If we isolate b from the equation 2, we get:

-0.2 = -2.5 (b - 0.4 ( 1.2 - 1.5a) )

(-0.2/-2.5) = b - 0.4 ( 1.2 - 1.5a)

0.08 = b - 0.48 + 0.6a

0.08 + 0.48 - 0.6a = b

0.56 - 0.6a = b

Then, replacing this on eq. 1 and solving for a, we get:

0.6 = 1.5 ( a + c ( b + 0.8 ) )

0.6/1.5 = a + cb + 0.8c

0.4 = a + c(0.56 - 0.6a) + 0.8c

0.4 = a + 0.56c - 0.6ca + 0.8c

0.4 - 0.8c - 0.56c = a(1 - 0.6c)

0.4 - 1.36c = a (1-0.6c)

Now, we have an expression for a in terms of c. Therefore, the value of a has no solutions if the denominator of the expression is equal to zero or if 1-0.6c is equal to zero. Then, the value of c for which there is no solutions (a,b) is calculated as:

1-0.6c = 0

1 = 0.6c

1/0.6 = c

5/3 = c

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.