Benchmark are numbers that are used as standards to which the rest of the data is compared to. When counting numbers using a number line, the benchmark numbers are the intervals written on the axis. For benchmark numbers of 10, the number line on top of the attached picture is shown. Starting from 170, the tick marks are added by 10, such that the next numbers are 180, 190, 200, and so on and so forth. When you want to find 410, just find the benchmark number 410.
The same applies to benchmark numbers in intervals of 100. If you want to find 170, used the benchmark numbers 100 and 200. Then, you estimate at which point represents 170. For 410, you base on the benchmark numbers 400 and 500.
Answer:
a) Adding -5x on both sides of the equation to remove the smaller x-coefficient
b) Adding -4 on both sides will remove the constant from the right side of the equation
Step-by-step explanation:
Given equation:
5x + (−2) = 6x + 4
a) What tiles need to be added to both sides to remove the smaller x-coefficient?
Smaller x-coefficient is 5x to remove the smaller x-coefficient
So, Adding -5x on both sides of the equation to remove the smaller x-coefficient
b) What tiles need to be added to both sides to remove the constant from the right side of the equation?
the constant on right side is 4
Adding -4 on both sides will remove the constant from the right side of the equation
Answer:
<h2>y=25p-45</h2>
Step-by-step explanation:
We first of all start by cumulating all the time she spent for break
lunch= 30min
break= 15min
total break= 45min
also the time taken to assemble one model plane 25min
let the number of hours be p
and the total number of model plane be y
The model is
y=25p-45
I don't know what 10p is so before I answer I'd have to know what that is
Over time, compound interest at any rate will outperform simple interest. When the rates are nearly equal to start with, compound interest will be greater in very short order. Here, it takes less than 1 year for compound interest to give a larger account balance.
In 30 years, the simple interest will be
... I = P·r·t = 12,000·0.07·30 = 25,200
In 30 years, the compound interest will be
... I = P·(e^(rt) -1) = 12,000·(e^(.068·30) -1) ≈ 80,287.31
_____
6.8% compounded continuously results in more total interest
