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GuDViN [60]
3 years ago
7

Determine the quotient: 2 4 over 7 ÷ 1 3 over 6.

Mathematics
2 answers:
bulgar [2K]3 years ago
6 0

Answer:

144/91

Step-by-step explanation:

24/7 ÷ 13/6

Keep the first fraction the same, change the sign, and flip the last fraction to solve

24/7 × 6/13 = 144/91

Levart [38]3 years ago
6 0

Answer:

144/91

Step-by-step explanation:

24/7 ÷ 13/6

Keep the first fraction the same, change the '÷' sign to '× ' and reciprocal of the last fraction to solve.  

24/7 × 6/13 = 144/91

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2.125

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IS THERE ANYONE GOOD AT MATH AND FINDING ROOTS ON GRAPHS THAT CAN HELP ME PLZ ??????
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Step-by-step explanation:

A root is a value for which a given function equals zero. When that function is plotted on a graph, the roots are points where the function crosses the x-axis. For a function, f(x) , the roots are the values of x for which f(x)=0 f ( x ) = 0 .

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fenix001 [56]

Answer: A

Step-by-step explanation: once you line the numbers up in order from least the greatest, the two middle numbers will be 12. Add 12 + 12 and you get 24. Then divide it by 2 and get 12. That is your median. Your 1st quartile will be 10. Your second quartile will be 15. Your minimum number is 4 and your maximum number is 18.

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3 years ago
The general term for the sequence 3/4,6/5,9/6,12/7,15/8
Ronch [10]

Answer:

B . \frac{3 n}{n+3}

Step-by-step explanation:

<u>Arithmetic sequence</u>:-

Step1:

Given sequence is \frac{3}{4} ,\frac{6}{5} ,\frac{9}{6} ,\frac{12}{7} ,\frac{15}{8}

Above sequence taking numerator terms

3,6,9,12,15 are in arithmetic progression

First term a= 3 and difference d = 3

now n t h term of given sequence is

t_{n} =a+(n-1)d

now substitute a =3 and d=3

t_{n} = 3+(n-1)3 = 3+3 n-3=3n

Step2:-

Above sequence taking denominator

4,5,6,7,8...

here a=4 and d=1

t_{n} =a+(n-1)d

t_{n} =4+(n-1)1

t_{n}=3+n

Final answer is the general term of given sequence is

t_{n} = \frac{3n}{3+n}

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4 years ago
What is the conjugate of 7-8i?
Serga [27]

Answer:

7 + 8i

Step-by-step explanation:

Conjugate of 7 - 8i is 7 + 8i

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