I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
Sorry I can't solve it.
Firstly, the equation is wrong. It is supposed to be <span>81 ≠ </span><span>30.25
Second of all, this is the best I got </span>81= <span><span>−<span>76 <span>(<span>cos<span>(r) </span></span></span></span></span>+ <span>106.25</span></span>
Answer:
4
Step-by-step explanation:
The factors of 4 are: 1, 2, 4
The factors of 12 are: 1, 2, 3, 4, 6, 12
Then the greatest common factor is 4.
Hope this helps! :D
Please mark brainlist
Answer: m1 = 0.5
M2= 0.375
M3 = 2.25
M4 = 3.0
M5 = 13.5
M6 = 0.285
M7 = 7.0
Step-by-step explanation:
(20,40)-(15,30)=(5,10), m1=5/10=0.5
(18,48)-(12,32)=(6,16), m2=3/8=0.375
(72,32)-(27,12)=(45,20), m3=9/4=2.25
(60,20)-(45,15)=(15,5), m4=3.0
(243,18)-(27,2)=(216,16), m5=27/2=13.5
(24,84)-(18,63)=(6,21), m6=2/7=0.285...
(84,12)-(63,9)=(21,3), m7=7.0
