Q4: a recipe calls for 3 cups of cornstarch. one cup of cornstarch weighs 4.5 ounces. how
much would 1 cup cost if the price of cornstarch was $0.98 per pound?

ANSWER:  

One cup of cornstarch weighs 4.5 ounces.The cost of 1 cup is $0.286 approximately.

SOLUTION:

Given, One cup of cornstarch weighs 4.5 ounces.  

The price of cornstarch was $0.98 per pound.

We need to find 1 cup cost.

Cost of a cup is product of cornstarch weight in cup and price of cornstarch per pound.

1 cup cost = cornstarch weight price of cornstarch per pound.

Here, both the terms are in different units, we need to convert them into single unit.

Let us convert ounces into pounds

We know that, 1 pound = 16 ounces

1 ounce = pound

Now, one cup of cornstarch weighs 4.5 ounces

Then, cup of cornstarch weight in pounds =

=

= 0.28125 pounds

Now, cost of 1 cup = cornstarch weight x price of cornstarch per pound

= 0.28125 x 0.98

= 0.275625

Hence, the cost of 1 cup is $0.286 approximately

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.