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laiz [17]
3 years ago
10

4# is a two digit number ,where # represents the digit at ones place . 7999 ÷ 4# lies between ---------

Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

7999 \div 4\# lies between 163.245 and 199.975

Step-by-step explanation:

Given

2 digit = 4#

Required

The range of 7999 \div 4\#

Let

\# = 0 --- the smallest possible value of #

So:

7999 \div 40 = 199.975

Let

\# = 9 --- the largest possible value of #

So:

7999 \div 49 = 163.245

<em>Hence, </em>7999 \div 4\#<em> lies between 163.245 and 199.975</em>

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Find LCM by division method<br><br> (i) 20,300 (ii) 1,05,22070
lisabon 2012 [21]

The LCM of 20 and 300 is 300.

<h3>How to find the LCM?</h3>

The multiple of 20 will be:

20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300

The multiple of 300 will be:

300 = 300, 600, 900

Therefore, the lowest common multiple of 20 and 300 is 300.

Learn more about LCM on:

brainly.com/question/18084619

#SPJ1

8 0
2 years ago
What is the numerical sum of angle DEA (X+25) and angle AEB (9x+10) ?
Sonbull [250]
10x + 35
If it is just the sum then it is this.
8 0
3 years ago
Becky spent 3/4 of an hour practicing the piano.She spent 1/2 of that time practicing her recital piece. What fractional part of
Ber [7]

Answer:

3/8

Step-by-step explanation:

First,

convert  3/4 of an hour to minutes;

If 1 hr = 60 mins

then 3/4 of 1hr = \frac{3}{4} *60\\ \\ =45 mins.Next, If 1/2 of 3/4 was spent practicing recital piece, it means that it is half of 45 mins;1/2 *45 = 22.5 minsLastly, convert the 22.5 mins into a fraction of an hour;22.5/60 = [tex]\frac{3}{8}

3 0
3 years ago
HELPP PLSSMFCODSM MATH TEST !
tresset_1 [31]

Answer:

17= addition

18 = 27

Let's solve your equation step-by-step.

43x−4=5−x

Step 1: Simplify both sides of the equation.

43x−4=5−x

43x+−4=5+−x

43x−4=−x+5

Step 2: Add x to both sides.

43x−4+x=−x+5+x

73x−4=5

Step 3: Add 4 to both sides.

73x−4+4=5+4

73x=9

Step 4: Multiply both sides by 3/7.

(37)*(73x)=(37)*(9)

x=277

Answer:

x=277

6 0
3 years ago
Suppose that you want to design a cylinder with the same volume as a given cylinder but you want to use a diff radius and height
Anton [14]

Answer:

Yes, r and h can be changed to produce same volume by;

Increasing "r" and decreasing "h" or by decreasing "r" and increasing "h".

Step-by-step explanation:

What the question is simply asking is if we can get same volume of a particular cylinder if we change the radius and height.

Now, volume of a cylinder is;

V = πr²h

Now, for the volume to remain the same, if we increase "r", it means we have to decrease "h", likewise, if we decrease "r", we now have to increase "h".

5 0
3 years ago
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