a triangle has a base of 15 inches and an area of 60 square inches. what is the height of the triangle? in your final answer, include all of your calculations. need learning how to do problem

3

step-by-step explanation:

in standard form:

a= dictionaries

b= maps

you have to buy 3 dictionaries, so put a 3 next to the a (dictionaries) so that you can multiply them by each other.

a3+b= 60 (the amount of money you have)

we already know the cost of a=15 so replace a with $15

($15 x 3)+b=60

multiply 15 and 3 to get $45 and replace it with $45

45+b=60

and then replace b with 15 to get the total:

45+15=60

that means you have $15 left after buying 3 dictionaries.

15 divided by the cost of a map ($5) equals 3.

so you can buy 3 maps.

sorry if it was confusing!

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)

d. 145*2 is 290 than the total angle is 360 then minus 290 then divide by two

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.