No, the trailer cannot hold the weight of the bricks. It is beyond the 900kg capacity of the trailer. The total weight of the bricks is 1,013.77 kilograms. The total weight was derived from getting the volume of the brick (0.051m x 0.102m x 0.203m), then multiplying the volume to the density of each brick (1.056 x 10^3m^3 x 1920kg/m^3). The weight of each brick is 2.03kg. Lastly, multiply the total number of bricks to the weight of each brick to get the total weight.
To determine the cost of each item, we need to set up equations. From the problem statement, we have three unknowns so we need three equations. We set up equations as follows:
let x cost of small pizzas
y cost of soda
z cost of salad
two small pizzas, a liter of soda, and a salad cost $14
2x + y + z = 14
one small pizza, a liter of soda, and three salads cost $15
x + y + 3z = 15
three small pizzas, a liter of soda, and two salads cost $22
3x + y + 2z = 22
Solving for x, y and z, we will have:
x = $ 5
y = $ 1
z = $ 3
To answer this kind of question, you need to insert the first equation into second equation or vice versa. The equation from the question should be:
1. 3 paperback books + 5 hardcover book = $80.10
2. 7 paperback books + 4 hardcover book= $100.65
It would be easier to convert the 1st equation into
5 hardcover book = $80.10- 3 paperback books
hardcover book= $16.02- 0.6 paperback books
Then insert it to 2nd equation:
7 paperback books + 4 ($16.02- 0.6 paperback books)= $100.65
7 paperback books + $64.08- 2.4 paperback books= $100.65
4.6 paperback books = $100.65- $64.08= $36.57
paperback book= $7.95
Insert paperback book price to the first equation to find the hardcover book price
3 paperback books + 5 hardcover book = $80.10
3 ($7.95)+ 5 hardcover book = $80.10
5 hardcover book = $80.10 - $23.85= $56.25
hardcover book = $11.25
Then one paperback and one hardcoverbook= $7.95 + $11.25= $19.20
Answer:
We have to determine the graph of the given inequality as:
on solving this equation we get:
Since we know that:
Now we know that when 
the value of:
Also for the value other than -1 the function being a quadratic function will always give a non-zero positive value.
Hence, range of the function in intervals could be written as:
(-∞,-1)∪(-1,∞).
Hence, the graph of the function will be the whole of the number line with a open circle on -1.
