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bogdanovich [222]
3 years ago
15

How do I teach my 4th grader to divide with two digits. Example 672/32 what are the steps?

Mathematics
2 answers:
Elanso [62]3 years ago
7 0
Answer/Work:
672 divided by 32
is 21 with remainder 0
= 21 R 0
= 21 0/32

Show Work:
(There are a few files that should help you see the steps of this problem, and examples (of)
 and how to do it) I hope this helps! If not, please let me know. :)


  <span><span> </span></span>

kipiarov [429]3 years ago
5 0
Ok so it's simple this is the way i learned it.
     ____
32 |672  first see how many times 32 goes into 6 which is 0 so put a 0 above 6

     _0___
32 |672   now you take the remainder from the last one which is 6 and combine it with the 7 so then you see how many times how many times 32 goes into 67 which is 2 so put a 2 over it.
     _02__
32 |672    next once again take the remainder which is 1 and combine it with the next number which is 2 to make 12 then find out how many times that goes into 32 

you basically do that till you get a number without a remainder and when you run out of numbers just put a decimal and zeros after it
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3 years ago
The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

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