The sides of the rectangla are its length and width. We know the length is 5m longer than the width. Since we dont know the width, we can say that the width is x and the length is x+5.
The perimeter, as we know, is equal to the sum of all sides. We have two “length” sides and two “width” sides, so our equation is:
x + x + x+5 + x+5 = 46
Lets add all of this:
4x + 10 = 46
4x = 46-10
4x = 36
x = 36/4
x = 9
So our width is 9m and the length is 14m.
The area of a rectangle is width times length, so:
9*14=126m^2.
Woo-Jin and Kiran were asked to find an explicit formula for the sequence 64\,,\,16\,,\,4\,,\,1,...64,16,4,1,...64, comma, 16, c
olya-2409 [2.1K]
Answer:

Step-by-step explanation:
The given sequence is
64, 16, 4, 1



It is a geometric series because it has a common ratio
.
First term is 64.
The explicit formula of a geometric series is

where, a is first term and r is common ratio.
Substitute a=64 and r=1/4 in the above function.

Therefore, the required explicit formula is
.
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
Use the lengths of corresponding sides
so answer is 16:40 which simplifies to 2:5.
Answer:
Step-by-step explanation:
to find out t, we divide by 3 both sides
-2/3=t
t=-2/3