The nickels in the envelope is $0. 12
<h3>How to determine the number</h3>
From the information given, we have that;
Dimes + nickels = $1. 20
1/4 dimes + dimes = $ 1. 35
Let dimes = d
Nickels = n
d + n = 1. 20 equation a
d/4 + d = 1. 35 equation b
Make 'd' subject from equation a
d = 1. 20 - n
Substitute into equation b
1. 20 - n/ 4 + 1. 20 - n = 1. 35
0. 3 - 0. 25n + 1. 20 - n = 1. 35
collect like terms
- 1. 25n = 1. 35 - 1. 5
- 1. 25 = - 0. 15
n = -0. 15/ -1. 25
n = 0. 12
Thus, the number of nickels in the envelope is $0. 12
Learn more about algebraic expressions here:
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Answer:
see below
Step-by-step explanation:
The equation of the hyperbola can be written as ...
((x -h)/a)² -((y -k)/b)² = 1
This has asymptotes ...
(x -h)/a ± (y -k)/b = 0
Solving for y, we have ...
y = ±(b/a)(x -h) +k
Filling in the given values a=6, b=8, h=1, k=2, we have ...
y = ±8/6(x -1) +2

Answer:
49 minutes
Step-by-step explanation:
To find the average time the machine was down per day, we need to add up the total time taken and then divide by the number of days. Since there are 5 days, we have:

= 49
So, the average time per day is 49 minutes.
Hope this helped! :)
Answer: LAST OPTION.
Step-by-step explanation:
To solve this exercise is important to remember the following definitions:
- The sum is the result obtained when we solve an addition.
- The product is the result obtained when we solve a mutiplication.
- Given a multiplication
, "a" and "b" are factors and "c" is the product.
In this case, you have this expression:

Notice that the there is an addition of two terms (
and
) inside the parentheses.
Outside the parentheses you can notice that the number
is multiplying the sum of the terms mentioned before.
Therefore, you can conclude that the best description for this expression is:
<em>The product of a constant factor of seven and a factor with the sum of two terms.</em>
Check the picture below.
let's notice that the base of the pyramid is triangle with a base of 5 and an altitude of 4, so it has an area of (1/2)(5)(4).
![\bf \textit{volume of a pyramid}\\\\ V=\cfrac{1}{3}Bh~~ \begin{cases} B=\stackrel{\textit{area of its}}{base}\\ h=height\\[-0.5em] \hrulefill\\ B=\frac{1}{2}(5)(4)\\ h=8 \end{cases}\implies V=\cfrac{1}{3}\left( \cfrac{1}{2}(5)(4) \right)(8)\implies V=\cfrac{1}{3}(10)(8) \\\\\\ V=\cfrac{80}{3}\implies V=26\frac{2}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20pyramid%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B1%7D%7B3%7DBh~~%20%5Cbegin%7Bcases%7D%20B%3D%5Cstackrel%7B%5Ctextit%7Barea%20of%20its%7D%7D%7Bbase%7D%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20B%3D%5Cfrac%7B1%7D%7B2%7D%285%29%284%29%5C%5C%20h%3D8%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B1%7D%7B3%7D%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%285%29%284%29%20%5Cright%29%288%29%5Cimplies%20V%3D%5Ccfrac%7B1%7D%7B3%7D%2810%29%288%29%20%5C%5C%5C%5C%5C%5C%20V%3D%5Ccfrac%7B80%7D%7B3%7D%5Cimplies%20V%3D26%5Cfrac%7B2%7D%7B3%7D)