Your answer is going to be B.
Let's go through each answer choice one at a time.
--------------------------------------------------
Choice A
3/4 < 5/7
3*7 < 4*5 ... cross multiply
21 < 20
The last inequality is false, so the original inequality is false
Choice A is false
--------------------------------------------------
Choice B
2/3 > 5/6
2*6 > 3*5 ... cross multiply
12 > 15
The last inequality is false, so the original inequality is false
Choice B is false
--------------------------------------------------
Choice C
5/8 > 6/10
5*10 > 8*6 ... cross multiply
50 > 48
The last inequality is true, so the original inequality is true
Choice C is true
--------------------------------------------------
Choice D
4/5 < 2/9
4*9 < 5*2 ... cross multiply
36 < 10
The last inequality is false, so the original inequality is false
Choice D is false
--------------------------------------------------
In summary, we have
A. False
B. False
C. True
D. False
So the final answer is choice C
Answer:
482190
Step-by-step explanation:
Answer:
let us consider a positive integer a
divided the positive integer a by 3, and let r be the remainder and B be the quotient and such that.
<h3>a=3b+r.........(1)</h3>
where =0,1,2,3
case 1:consider r=0
equation (1)becomes
<h3 /><h3>a=3b</h3>
on squaring both the side
<h3>a²=3m</h3>
- where m=3b²
- case 2: Let R=1
<h3>a=3b+1</h3>
- squaring on both the side we get
- a²=(3b+1)²
- a²=(3b) ²+1+2x(3b)x1
- a²=9b²+6b+1
- a²=3(3b²+2b) +1
<h3 /><h3>a²=3m+1</h3>
- where m=3b²+2b
- case3:let r=2
- equation (1)becomes
<h3>a=3b+2</h3>
- squaring on both the sides we get
- a²(3b+2)²
- a²=9b²+4(12x3bx2
- a²9b²+(12b+3+1)
- a²3(3b²+4b+1)+1
<h3 /><h3>a²3m+1</h3>
where m=3b²+4b+1
: square of any positive integer is of the form 3M or 3M + 1
: hence proved.
