Anna buys bananas, grapes, and apples. she buys 5 times as many bananas as grapes. she buys 2 more pounds of apples than bananas. altogether, anna buys 35 pounds of fruit. how many pounds of each fruit did she buy?

Anna Bought 3 pounds of grapes, 15 pounds of bananas and 17 pounds of apples.

Step-by-step explanation:

Given:

Anna buys bananas, grapes, and apples.

She buys 5 times as many bananas as grapes.

Let Amount of grapes be 'x' pounds.

So According to question given;

Amount of bananas =

Also Given:

She buys 2 more pounds of apples than bananas.

It means that Amount of apples is 2 more than amount of bananas.

So Amount of apples = 2 + amount of bananas.

But Amount of bananas =

So Amount of apples =

Now Given:

Total Amount of fruit she bought = 34 pounds.

We need to find amount of each fruit did she buy.

Now Total Amount of fruit she bought is sum of Amount of grapes and Amount of bananas and Amount of apples.

framing in equation form we get;

Amount of grapes = 3 pounds

Amount of bananas =

Amount of apples =

Hence Anna Bought 3 pounds of grapes, 15 pounds of bananas and 17 pounds of apples.

7.3

step-by-step explanation:

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.