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lana [24]
3 years ago
14

Whats 600 divided by 4

Mathematics
2 answers:
astra-53 [7]3 years ago
7 0
The answer is 180 my teacher showed a trick called the house and in ur answer theirs no remainder it's just 180
Taya2010 [7]3 years ago
6 0
4 goes into six hundred 150 times. So the answer is 150.
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The following table gives annual life insurance premiums per $1,000 of face value. Use the table to determine the annual premium
Papessa [141]
The answer would be C !
4 0
2 years ago
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elena-s [515]

Answer:

A. (0,1) and (1,2)

Step-by-step explanation:

I graphed it

6 0
3 years ago
Answer correclty PLZZZ
ioda

Answer: option C is the correct answer

Step-by-step explanation:

The system of linear equations is

10x + 7y = 12 - - - - - - - 1

8x + 7y = 18 - - - - - - - 2

Since the coefficient of y is the same in equation 1 and equation 2, we will eliminate y by subtracting equation 2 from equation 1, it becomes

10x - 8x + 7y - 7y = 12 - 18

2x = -6

x = - 6/2 = - 3

Substituting x = - 3 into equation 1, it becomes

10×-3 + 7y = 12

-30 + 7y = 12

Let the constants be on the right hand side and the term containing y be on the left hand side. It becomes

7y = 12 + 30

7y = 42

y = 42/7

y = 6

C) (−3, 6)

6 0
3 years ago
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Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
Answer this pls<br>don't send links​
Irina18 [472]

Answer:

a 20 times

Step-by-step explanation:

5 0
3 years ago
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