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vovangra [49]
3 years ago
10

Im just looking for 25/4 (that's a fraction) divided in 3.

Mathematics
2 answers:
Hatshy [7]3 years ago
8 0

Answer:

\frac{25}{12}

Step-by-step explanation:

1) Use this rule: a ÷ b/c = a × c/b.

\frac{25}{4}  \times  \frac{1}{3}

2) Use this rule: a/b × c/d = ac/bd.

\frac{25 \times 1}{4 \times 3}

3) Simplify 25 × 1 to 25.

\frac{25}{3 \times 4}

4) Simplify 3 × 4 to 12.

\frac{25}{12}

<em><u>Therefor</u></em><em><u>,</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>25</u></em><em><u>/</u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>decimal</u></em><em><u> </u></em><em><u>form</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>would</u></em><em><u> </u></em><em><u>be</u></em><em><u>,</u></em><em><u> </u></em><em><u>2.08</u></em><em><u>3</u></em><em><u>3</u></em><em><u>3</u></em><em><u>3</u></em>.

ANTONII [103]3 years ago
5 0
Answer: 2.083
Have a good day :D
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Step-by-step explanation:

8 0
3 years ago
How many positive integers between 1000 and 9999 inclusive
bekas [8.4K]

Answer:

Step-by-step explanation:

a.

first number is  1000-1+9=1008

9)1000(1

    9

-------

     10

       9

    -----

       10

         9

       ----

         1

       ----

last number is 9999

9| 9999

  ---------

    1111 |0

    --------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

  ---------

  |4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total  numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

  ---------

  |  142-6

1000-6+7=1001

7 | 9999

  |---------

    1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

        70

       ----

        300

        280

        ------

           20

           ---

1000-20+35=1015

35)9999(285

     70

    ----

     299

     280

     -----

        199

        175

        ----

          24

         ----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

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What is the result of the following division?
Vlad1618 [11]

For this case we must indicate the result of the following expression:

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Answer:

See the attached image

Option A

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