Width = W
Length = 12W - 1
Perimeter = 2L + 2W = 37
In the perimeter equation substitute the length equation in for L to get the equation in terms of W
2L + 2W = 37
2(12W - 1) + 2w = 37
Distribute
2(12W - 1) + 2w = 37
24W - 2 + 2w = 37
26W = 39
W = 3/2 = 1.5
Width = 1.5
Lastly, solve for length
L = 12W - 1
L = (12 • 1.5) - 1
L = 18 - 1
L = 17
Length = 17
Answer:
approximately 4.978%
Step-by-step explanation:
The probability of both happening is 3/13*11/51, or 33/663, or 11/221 (approximately 4.978%). Two cards are drawn one after the other with replacement from a pack of 52 cards.
Step-by-step explanation:
2 right triangles are formed.
x is the angle btw wire and grnd
tan x = 10/5 = 2
tanx = ht of tower/(5+14)
2 = tower ht/19
tower ht = 38
length of wire =
Answer:
-4
Step-by-step explanation:
Subtract 1 - 5 = -4...
Since 1 is less than 5, we know that the difference is negative, so we can rewrite this expression as -(5-1).
5 - 1 = 4, so - (5-1) = -4.
<h3>
Answer:</h3>
System
Solution
- p = m = 5 — 5 lb peanuts and 5 lb mixture
<h3>
Step-by-step explanation:</h3>
(a) Generally, the equations of interest are one that models the total amount of mixture, and one that models the amount of one of the constituents (or the ratio of constituents). Here, there are two constituents and we are given the desired ratio, so three different equations are possible describing the constituents of the mix.
For the total amount of mix:
... p + m = 10
For the quantity of peanuts in the mix:
... p + 0.2m = 0.6·10
For the quantity of almonds in the mix:
... 0.8m = 0.4·10
For the ratio of peanuts to almonds:
... (p +0.2m)/(0.8m) = 0.60/0.40
Any two (2) of these four (4) equations will serve as a system of equations that can be used to solve for the desired quantities. I like the third one because it is a "one-step" equation.
So, your system of equations could be ...
___
(b) Dividing the second equation by 0.8 gives
... m = 5
Using the first equation to find p, we have ...
... p + 5 = 10
... p = 5
5 lb of peanuts and 5 lb of mixture are required.
