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mariarad [96]
4 years ago
14

Please help what’s the missing length of TU?

Mathematics
1 answer:
svetlana [45]3 years ago
3 0

Answer:

its b

Step-by-step explanation:

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I need HELP!! With <br> 31, 33, 35 please explain
rusak2 [61]
For 31 you need 6% OF 50 meaning you need to multiply. 50*0.06=3. The answer for 31 is 3.

To find 33 you need to divide 1 by 0.4 to find how much you need to multiply 28 by. 1/0.4= 2.5 then you need to do 2.5*28 which would equal 70. The answer for 33 is 70.

To find 35 you need to notice that it DECREASES in meaning the bigger number (20) would be the denominator of the fraction and 15 would be the numerator. Therefore the fraction would b 15/20 or 75% of 20 would equal out to 15.
3 0
3 years ago
Read 2 more answers
At the north campus of a performing arts schools, 10% of the students are music majors. At the south campus, 70% of the students
ad-work [718]

Answer:

10000

Step-by-step explanation:

3 0
3 years ago
-(6x+6)+2=-6x-3 please solve
makkiz [27]

Answer:

There is no solution.

Step-by-step explanation:

To solve, use the order of operations. Follow them until you solve for x or just numbers remain.

-(6x + 6) + 2 = -6x - 3

-6x - 6 + 2 = -6x - 3

-6x - 4 = -6x - 3

-4 = -3

The answer is no solution since this is a false statement.

3 0
3 years ago
-7 2/3 + (-5 1/2) + 8 3/4=
myrzilka [38]
-4.41666666667



Here u go
3 0
2 years ago
Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible
alisha [4.7K]

Answer:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999

6 0
3 years ago
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