Answer:
He must charge minimum £0.84 for one hot dog.
Step-by-step explanation:
Ingredients cost for hot dogs at the supermarket,
Price of dog sausages tin of 8 = 50p ≈ £0.5
Price of dog bun packs of 6 = 90p ≈ £0.90
Number of tins required for 150 hot dogs = 
= 6.25
So student will purchase 7 tins.
Cost of 7 tins = 7 × 0.5
= £3.5
Number of dog buns packets required = 
= 25
Cost of 25 packs = 25 × 0.9
= £22.5
Total cost of the material used for 150 buns = 22.5 + 3.5
= £26.0
Let the selling price of each hot dog = £x
Then selling price of 150 hot dogs = £150x
Students wants the profit = £100
So, Selling price - Cost price = profit
150x - 26 = 100
150x = 126
x = 
x = £0.84
Therefore, he must charge minimum £0.84 for one hot dog.
The length of the envelope is 6.3 inches
Step-by-step explanation:
Width of the envelope = 3 inches
Diagonal of the envelope = 7 inches
To find:
The length of the envelope.
Let the length of the envelope be 'l'
We can use pythogoras theorem to calculate the length.
l = √(49-9)
l = √(40)
l = 6.3 inches
The length of the envelope is 6.3 inches
y = 3x - 1
x - y = -9
Subs y in second equation.
x - (3x - 1) = -9
x - 3x + 1 = -9
-2x = -9 - 1
-2x = -10
x = -10/-2
x = 5
Subs x
y = 3x - 1
y = 3.5 - 1
y = 15 - 1
y = 14
Answer:
C₂₃ = -186
↓
C₁₃ = -32
↓
C₃₁ = 6
↓
C₁₁ = 27
↓
C₂₁ = 28
↓
C₃₃ = 38
↓
C₂₂ = 56
↓
C₃₂ = 90
↓
C₁₂ = 115
Step-by-step explanation:
The given matrices are;
![A = \left[\begin{array}{ccc}1&7&-1\\5&-2&-9\\-3&8&3\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%267%26-1%5C%5C5%26-2%26-9%5C%5C-3%268%263%5Cend%7Barray%7D%5Cright%5D)
![B = \left[\begin{array}{ccc}5&1&7\\3&15&-2\\-1&-9&25\end{array}\right]](https://tex.z-dn.net/?f=B%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%267%5C%5C3%2615%26-2%5C%5C-1%26-9%2625%5Cend%7Barray%7D%5Cright%5D)
The cross product of the matrices is found as follows;
![A \cdot B = \left[\begin{array}{ccc}1&7&-1\\5&-2&-9\\-3&8&3\end{array}\right] \times \left[\begin{array}{ccc}5&1&7\\3&15&-2\\-1&-9&25\end{array}\right]](https://tex.z-dn.net/?f=A%20%5Ccdot%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%267%26-1%5C%5C5%26-2%26-9%5C%5C-3%268%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%267%5C%5C3%2615%26-2%5C%5C-1%26-9%2625%5Cend%7Barray%7D%5Cright%5D)
C₁₁ = 1×5 + 7×3 + (-1) × (-1) = 27
C₁₂ = 1×1 + 7×15 + (-1)×(-9) = 115
C₁₃ = 1×7 + 7×(-2) + (-1)×25 = -32
C₂₁ = 5×5 + (-2)×3 + (-9) × (-1) = 28
C₂₂ = 5×1 + (-2)×15 + (-9)×(-9) = 56
C₂₃ = 5×7 + (-2)×(-2) + (-9)×25 = -186
C₃₁ = (-3)×5 + 8×3 + 3 × (-1) = 6
C₃₂ = (-3)×1 + 8×15 + 3×(-9) = 90
C₃₃ = (-3)×7 + 8×(-2) + 3×25 = 38
Therefore, we get;
![A \cdot B = \left[\begin{array}{ccc}1&7&-1\\5&-2&-9\\-3&8&3\end{array}\right] \times \left[\begin{array}{ccc}5&1&7\\3&15&-2\\-1&-9&25\end{array}\right] = \left[\begin{array}{ccc}27&115&-32\\28&56&-186\\6&90&38\end{array}\right]](https://tex.z-dn.net/?f=A%20%5Ccdot%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%267%26-1%5C%5C5%26-2%26-9%5C%5C-3%268%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%261%267%5C%5C3%2615%26-2%5C%5C-1%26-9%2625%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D27%26115%26-32%5C%5C28%2656%26-186%5C%5C6%2690%2638%5Cend%7Barray%7D%5Cright%5D)
In increasing order, we have;
C₂₃ = -186
↓
C₁₃ = -32
↓
C₃₁ = 6
↓
C₁₁ = 27
↓
C₂₁ = 28
↓
C₃₃ = 38
↓
C₂₂ = 56
↓
C₃₂ = 90
↓
C₁₂ = 115
Answer:
7 - z = 0
- z + 7 = 0
- z = - 7
z = 7
Thus, The answer is 7
<u>-TheUnknownScientist</u>
