Step-by-step explanation:
Q7) 10% of 30=3
40% of 30 =3×4=12
30-12=18
the sale price of radio is 18
q8)purchase lose lemmons:0.99÷2=0.495
Bag of 6=2.50÷6=0.4166666667
so the better value is the bag of 6 for 2.50.
Answer:
Step-by-step explanation:
We must divide the number of pages by the number of minutes in order to find how much she would answer per minute.
So, our equation looks like this:
12 ÷ 2 1/2 = rate per minute
= 12 ÷ 5/2 (2 1/2 as an improper fraction) = 4.8 pages
Hope this helped!
t = 1.33 sec
Solution:
Given data:
Velocity
= 17.5 ft/s
Height
= 5 ft
The height can be modeled by a quadratic equation

where h is the height and t is the time


a = –16, b = 17.5, c = 5
It looks like a quadratic equation. we can solve it by quadratic formula,






Time cannot be in negative. So neglect t = –0.235.
t = 1.33 sec
Hence Amy have to react 1.33 sec before the volleyball hits the ground.
Answer:
6
Step-by-step explanation:
1 shook 2,3, and 4's hands
2 shook 3, and 4's hands
3 shook 4's hands
4 already got his hand shaken by everyone
1-2, 1-3, 1-4.
2-3, 2-4
3-4
(Anyone feel free to correct me. I'm not amazing at math just trying to help out.)
Answer:

Step-by-step explanation:
To find x₁ and x₂ :
![\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] \times \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%261%5C%5C5%264%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step 1
Multiply first 2 x 2 matrix with 2 x 1 vector, we get
![\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2B%20%20x_2%5C%5C5x_1%26%2B%20%204x_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 2
Add the 2 x 1 matrices on LHS, we get
![\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2Bx_2%26%2B11%5C%5C5x_1%26%2B4x_2%26-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 3,
we get

and

Step 4,
Simplify, we get

Step 5,
multiply eqn(1) by 4
we get

Step 6,
eqn (2) - eqn(3)
we get

substituting in eqn (1), we get

so, we get

Therefore
