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3 years ago

12

A Family purchased tickets to the movies in spent a total of $40.75. The family purchase five tickets there was a $.90 processi

ng fee for each ticket. How much was 1 ticket.

1 answer:

muminat3 years ago

7 0

Answer:

$7.25

Step-by-step explanation:

Total amount spent on the ticket = $40.75

Number of ticket purchased = 5

Processing fee per ticket = $0.90

Unknown:

Cost of each ticket = ?

Solution:

To solve this problem, we need to find the processing fee for the 5 tickets purchase;

Processing fee = number ticket x processing fee per ticket

Processing fee = 5 x 0.9 = $4.5

Now, the real cost a ticket = $40.75 - $4.5 = $36.25

So, cost per ticket is;

Cost = $\frac{36.25}{5}$ = $7.25

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Write and then evaluate an expression for the following number line

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Answer:

ok there are know boxes

Step-by-step explanation:

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3 years ago

Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

<em>So, the sides are: 38.5, 39.5, 40.5 and 41.5</em>

<em></em>

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

<em>So presently, Inzo is 15 years old and Miz is 10 years old</em>

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

5 0

2 years ago

Divide 54 into two parts so that four times the greater equals five time the less

LUCKY_DIMON [66]

Let the both the parts be 'x' and 'y'
x>y
We know that 4x = 5y
so x = 5y/4
x + y = 54
Replacing x we get
(5y/4) + y = 54
(5y +4y)/4 = 54
9y = 216
y = 24
Replacing y in ' x+y = 54'
we get x = 30

Hence the larger part is 30 whereas the smaller part is 24

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3 years ago

2x + y - z = -8

Andreyy89

Answer:

Multiply row 1 by $\frac{1}{2}$ .

Step-by-step explanation:

The augmented matrix of the system of linear equation is described below:

$\left[\begin{array}{cccc}2&1&-1&-8\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Where $a_{11} = 2$ , if we need to create $a_{11} = 1$ , we need to multiply row 1 by $\frac{1}{2}$ , that is to say:

$\left[\begin{array}{cccc}1&\frac{1}{2} &-\frac{1}{2} &-4\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Hence, the correct answer is: Multiply row 1 by $\frac{1}{2}$ .

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3 years ago

The length of a rectangle is x² + 6x + 3 and the width is 3x² + 4x - 2. Which expression represents the are of the rectangle

miss Akunina [59]

Answer:

Step-by-step explanation:

area of rectangle=length×width

=(x²+6x+3)×(3x²+4x-2)

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=3x^4+22x³+31x²-6

7 0

3 years ago