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3 years ago

3. Brenda's school is selling tickets to a spring musical. On the first day of

Mathematics

1 answer:

Natalija [7]3 years ago

4 0

Answer:

The Answer is that Senior Citizen Tickets cost: $4 and Child tickets cost: $7.

Step-by-step explanation:

Let s = the cost of senior citizen tickets

Let c = the cost of child tickets

The number of tickets sold for each type added together equals the sales for each day. Equations below:

Day 1

3s + 9c = $75

Solve for s:

3s = 75 - 9c

s = 25 - 3c

Day 2

8s + 5c = $67

By substitution:

8(25 - 3c) + 5c = 67

200 - 24c + 5c = 67

-19c = -133

c = -133 / -19 = $7 cost for child tickets.

Solve for s:

s = 25 - 3c

s = 25 - 3(7)

s = 25 - 21 = $4 cost for senior citizen tickets.

Proofs:

Day 1

3s + 9c = $75

3(4) + 9(7) = 75

12 + 63 = 75

75 = 75

Day 2

8s + 5c = $67

8(4) + 5(7) = 67

32 + 35 = 67

67 = 67

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Hey there!

$\large\boxed{10u^2}$

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Image attached.

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Makovka662 [10]

The value of 29 - |7 - 11| = 25

We have the following expression -

29 - |7 - 11|

We have to find its value.

<h3>Find the value of the expression -</h3><h3>X + | $\pi$ Y | where Y < 0</h3>

We have -

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Answer:

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3 years ago

Help please!!!!!!!!!

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ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

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3 years ago