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
Answer:
A
Step-by-step explanation:
Given the 2 equations
4x + 5y = - 12 → (1)
- 2x + 3y = - 16 → (2)
Eliminate the x- term by multiplying (2) by 2 and adding the result to (1)
- 4x + 6y = - 32 → (3)
Add (1) and (3) term by term
11y = - 44 ( divide both sides by 11 )
y = - 4
143 is composite. To be prime it has to have only 2 divisors i.e it's self and one but since it has 4 divisors ( 143, 1, 11 and 13) it is composite.
Answer:
$120
Step-by-step explanation:
Answer: (3.65, 3.74)
Step-by-step explanation:
Now, when we round a number to a given decimal place, we need to look at the decimal to the left of the decimal place where we are rounding.
if it is 5 or more, we round up
if it is smaller than 5, we round down.
Then for example, if we want to round:
4.534
to the first decimal place, we need to look at the second decimal place.
we can see that it is a 2.
Then we round down, and get:
4.5
Now we know that, when rounding to the first decimal place, we have:
x = 3.7
Then the maximum value that x could have is:
x = 3.74 (because here we round down to 3.7)
and the minimum value that x could have is:
x = 3.65 (because here we round up to 3.7)
Then the error interval for x is:
(3.65, 3.74)
