Please someone help me with these 5 questions , please. 1. Find the area of the triangle.

A. 154 ft2
B. 58 ft2
C. 77 ft2
D. 38.5 ft2

2. Find the area for the following figure.

A. 125.86 cm2
B. 62.93 cm2
C. 26.5 cm2
D. 137.45 cm2

3. Find the area of the parallelogram

A. 29.44 cm2
B. 8.7 cm2
C. 14.72 cm2
D. 7.36 cm2

4. Find the area for a triangle base 14 cm , height 5.5 cm

A. 19.5 cm2
B. 77. cm2
C. 38.5 cm2
D. 770 cm2

5. Find the area of a parallelogram base 5 cm, height 20 cm

A. 100 cm2
B. 90 cm2
C. 50 cm2
D. 25 cm2

answer: b)

step-by-step explanation:

its a dot plot because since the score is not high numbers.

- r3ktforgood ☕

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

1 1/7 hope i

step-by-step explanation:

8/7 = (1 * 7 + 1) / 7 = 1 + 1/7 = 1 1/7

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.