5(4)+6(5)^2-(4)^3
20+150-64
170-64=106
Answer:
$30
Step-by-step explanation:
Let 6 cans represent an "order" of fruit juice. 72 cans means 12 orders (72 divided by 6). Since each order costs $2.50, 12 orders costs $30 (12 multiplied by 2.5).
AX = b or ![\left[\begin{array}{cc}1&1\\0.1&0.05\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}15\\1.10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%261%5C%5C0.1%260.05%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D15%5C%5C1.10%5Cend%7Barray%7D%5Cright%5D)
<h3>
The above matrix can be used to determine the number of dimes x and nickels y.</h3>
Step-by-step explanation:
Here, total number of coins = 15
Let us assume the number of dimes = x
Also, let us assume the number of nickels = y
So, x + y = 15 .... (1)
Again, 1 dime = $0.1
So, x dimes = x ($0.1) = $ (0.1 x)
1 nickel = $0.05
So, y nickels = y ($0.05) = $ (0.05 y)
Also, total value of coins = $1.10
⇒ 0.1 x + 0.05 y = 1.10 .... (2)
So, from (1) and (2) the matrix can be written as:
AX = b or ![\left[\begin{array}{cc}1&1\\0.1&0.05\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}15\\1.10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%261%5C%5C0.1%260.05%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D15%5C%5C1.10%5Cend%7Barray%7D%5Cright%5D)
The above matrix can be used to determine the number of dimes x and nickels y.
Suppose you roll a red number cube and a blue number cube. What is the probability that you will roll a 5 on the red cube and a 1 or 2 on the blue cube?
<h3>SOLUTION : </h3>
The radius of the circle is 4 inch and has an arc length of 5.8 inches . We know that if the central angle is 360° then the arc length is
. Similarly when angle is theta ,
Now substitute respective values , from the given data ,

This is our required answer!