Answer:
The given expanded sum of the series is 
Step-by-step explanation:
Given problem can be written as

To find their sums:
Now expanding the series
That is put n=5,6,7,8,9 in the given summation
![\sum\limits_{n=5}^{9}3n+2=[3(5)+2]+[3(6)+2]+[3(7)+2]+[3(8)+2]+[3(9)+2]](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bn%3D5%7D%5E%7B9%7D3n%2B2%3D%5B3%285%29%2B2%5D%2B%5B3%286%29%2B2%5D%2B%5B3%287%29%2B2%5D%2B%5B3%288%29%2B2%5D%2B%5B3%289%29%2B2%5D)
![=[15+2]+[18+2]+[21+2]+[24+2]+[27+2]](https://tex.z-dn.net/?f=%3D%5B15%2B2%5D%2B%5B18%2B2%5D%2B%5B21%2B2%5D%2B%5B24%2B2%5D%2B%5B27%2B2%5D)
(adding the terms)

Therefore 
Therefore the given sum of the series is 
The given expanded sum of the series is 
Answer:
The inequality represents the scenario is
.
Step-by-step explanation:
Given:
Number of minutes used to make cookie written on box = 30 minutes.
Number of minutes used to make cookies = 
We need to write inequality represents the scenario
Solution:
We can say that;
Number of minutes used to make cookies should be less than or equal to Number of minutes used to make cookie written on box.
framing in equation form we get;

Hence The inequality represents the scenario is
.
Answer:
p ∈ IR - {6}
Step-by-step explanation:
The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2
is all R2 ⇔
And also u and v must be linearly independent.
In order to achieve the final condition, we can make a matrix that belongs to
using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.
Let's make the matrix :
![A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%26p%262%5Cend%7Barray%7D%5Cright%5D)
We used the first vector ''u'' as the first column of the matrix A
We used the second vector ''v'' as the second column of the matrix A
The determinant of the matrix ''A'' is

We need this determinant to be different to zero


The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that 
We can write : p ∈ IR - {6}
Notice that is
⇒


If we write
, the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.
Answer: £164.50
175 decreased by 6% = 164.5
Absolute change (actual difference):
164.5 - 175 = - 10.5
Step-by-step explanation:
175 - Percentage decrease =
175 - (6% × 175) =
175 - 6% × 175 =
(1 - 6%) × 175 =
(100% - 6%) × 175 =
94% × 175 =
94 ÷ 100 × 175 =
94 × 175 ÷ 100 =
16,450 ÷ 100 =
164.5
£164.50
Answer:
3.33
Step-by-step explanation: