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NeTakaya
3 years ago
11

6 times a number is greater than 20 more than that number what are the possible values of that number

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0
Solve 6x>x+20.
6x-x>20
5x>20
x>4

Answer: all values of x (including decimals and irrationals) greater than 4 will satisfy the given requirement that 6 time x is greater than x+20.
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for independence day celebration 25000 flags were distributed to 70 schools .if 40 shcool receive 250 each and the remainder dis
aleksandr82 [10.1K]

Answer:

500 flags each

Step-by-step explanation:

If 40 schools receive 250 each, then in total, that's 10,000

25,000 - 10,000 = 15,000.   There are 30 schools remaining, so we divide.  15,000 / 30 = 500, so...

40 schools get 250 flags each, and the other 30 schools get 500 flags each.

5 0
3 years ago
Factorise 36x³y-225xy³​
blagie [28]
<h3>hi</h3><h3>here is ur answer:</h3>

6 0
3 years ago
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This is Q2:5 Weekly Math Review
mezya [45]

Answer:

28.50+x or 28.50x

Step-by-step explanation:

6*4.75=28.50

6*x=6x or x

7 0
3 years ago
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Write the following decimal number in its equivalent fraction form. <br> -3.143
Rzqust [24]
-3.143 is equivalent to -(3143/1000)
8 0
3 years ago
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Find the area of this regular polygon.<br> Round to the nearest tenth.<br> 8.65 mm<br> [? ]mm2
nadezda [96]

Answer:

Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:

a=√{b²+c²−2bc×cos40°}

a=√{149.645−149.645cos40°}

Area Nonagon = (9/4)a²cos40°

=9/4[149.645−149.645cos40°]cot20°

=336.70125[1−cos(40°)]cot(20°)

Applying an identity for the cos(40°) does not get us very far…

= 336.70125[1−(cos2(20°)−1)]cot(20°)

= 336.70125[2−cos2(20°)]cot(20°)

= 336.70125[2−(1−sin2(20°))]cot(20°)

= 336.70125[1+sin2(20°)]cos(20°)sin(20°)

= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²

3 0
3 years ago
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