A rectangular garden has an area if
12a^2-5a-2m^2
A.) Write the area as the product of two binomials with integer coefficients
B.) the garden is to be completely enclosed by a path 1 m wide. find and simplify an expression for the area of the path
C.) the path is concrete, poured to a depth of 10 cm. calculate the volume in m^3 of concrete used if a=6
PLEASE help:) thanks:)
A.) To write the area as the product of two binomials with integer coefficients, we need to factor the given expression.
The area of the rectangular garden is given by 12a^2 - 5a - 2m^2.
To find the factors, we need to find two numbers that multiply to give -24m^2 and add up to -5a.
The factors of -24m^2 are (-1, 24m^2), (1, -24m^2), (-2, 12m^2), (2, -12m^2), (-3, 8m^2), (3, -8m^2), (-4, 6m^2), (4, -6m^2), (-6, 4m^2), (6, -4m^2), (-8, 3m^2), (8, -3m^2), (-12, 2m^2), (12, -2m^2), (-24m, 1), and (24m^2, -1).
The factors of -24m^2 that add up to -5a are (-8, 3m^2) or (8, -3m^2).
Therefore, we can write the area as the product of two binomials: (3m^2 - 8)(4m^2 + 1).
B.) To find the area of the path, we need to add the area of the garden to the area of the path. The path is 1m wide on all sides, so we need to add 1m to both the length and width of the garden.
The area of the path is given by (length + 2)(width + 2), which simplifies to (3m^2 + 6)(4m^2 + 3).
C.) To calculate the volume of concrete used, we need to multiply the area of the path by the depth of the concrete.
The volume of concrete used is given by (area of path)(depth of concrete).
If a = 6, then the volume of concrete used is ((3(6)^2 + 6)(4(6)^2 + 3))(10 cm).
Simplifying the expression, we have (3(36) + 6)(4(36) + 3)(10 cm).
Calculating further, we have (108 + 6)(144 + 3)(10 cm).
Simplifying the expression, we have (114)(147)(10 cm).
Finally, multiplying the values, we have 16,7580 cm^3.
A.) To write the area as the product of two binomials with integer coefficients, we need to factor the given expression.
12a^2 - 5a - 2m^2
To factor this expression, you'll need to look for two binomials that multiply to give you the desired expression.
Start by multiplying the coefficient of the squared term (12) by the constant term (-2m^2). The result is -24m^2.
Now, think of two numbers whose product is -24 and whose sum is -5 (the coefficient of the middle term). In this case, the numbers are -8 and 3.
Now, rewrite the middle term (-5a) as the sum of -8a and 3a.
12a^2 - 8a + 3a - 2m^2
Now, group the terms:
(12a^2 - 8a) + (3a - 2m^2)
Now, factor out the greatest common factor (GCF) from each group:
4a(3a - 2) + (3a - 2m^2)
Notice that both groups have (3a - 2) as a common factor. So, we can factor it out:
(4a + 1)(3a - 2)
Therefore, the area of the rectangular garden can be written as the product of two binomials with integer coefficients: (4a + 1)(3a - 2).
B.) To find the area of the path surrounding the garden, we need to add the extra width of the garden that is covered by the path to both the length and width of the garden.
Let's represent the length and width of the garden as L and W, respectively.
With a path of 1 m wide, the new length of the garden will be L + 2 and the new width will be W + 2.
The area of the garden with the path can be calculated by multiplying the new length and width:
(L + 2)(W + 2)
To simplify the expression, we need to expand it:
LW + 2L + 2W + 4
Therefore, the expression for the area of the path is LW + 2L + 2W + 4.
C.) To calculate the volume of concrete used for the path, we need to find the product of the area of the path and the depth of the concrete. The depth is given as 10 cm, which can be converted to meters by dividing by 100.
The volume of concrete used is given by the expression:
(LW + 2L + 2W + 4) * depth
Substituting the value of a = 6, we can calculate the concrete volume. However, as you haven't provided the values of L and W, the specific concrete volume cannot be determined.
But, once you have the values for L and W, you can substitute them into the expression and then multiply by the depth to find the volume of concrete used in cubic meters (m^3).
12a^2-5a-2m^2
= (3a -2m)(4a + m)
If we add 1 m to the end of each length and width
area of garden including path
= (3a-2m+2)(4a+m+2)
So area of path
= (3a-2m+2)(4a+m+2) - (3a -2m)(4a + m)
I will let you do the expanding and simplifying
sub in a = 6 into your answer for the path which will still contain the variable m
volume of concrete
= .1(answer you have above) m^3