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krok68 [10]
3 years ago
14

Lauren coordinates a construction projects for a cement company. A government project requires constructing two rectangular conc

rete slabs of dimensions 24× 24× 1 feet. Additionally, the company sends a 20% surplus of concrete to ensure the job can be completed. If a cement truck can carry a maximum of 8 cubic yards of cement, what's the fewest number of trucks that Lauren should send? A)1 B)2 C)3 D)4 E)5lar
Mathematics
1 answer:
MakcuM [25]3 years ago
6 0

The fewest number of trucks Lauren should send is D) 4 trucks.

Step-by-step explanation:

Step 1:

The rectangular slab's dimensions are 24 \times 24 \times 1 feet. Each truck can carry 8 cubic yards of cement.

First, we need to determine the volume of the slabs in yards. 1 foot = 0.333 yards. So 24 feet = 24\times 0.3333 = 8 yards.

The volume of the slab = 8 \times 8 \times 0.3333 = 21.3312 cubic yards.

Step 2:

The company sends a surplus of 20% to make sure the job can be completed. So the total cement sent is the required volume and an extra 20%.

The total cement sent = The required cement + 20%.

                                      = 21.3312 + 20% = 25.597 cubic yards.

Step 3:

So to find the number of trucks needed, we divide the cement sent by the load each truck can carry. Each truck can carry 8 cubic yards of cement. So

The number of trucks needed = \frac{therequiredload}{load per truck} = \frac{25.597}{8} = 3.199625.

If 3.199 trucks are needed, it means 4 trucks are needed which is option D.

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3 years ago
Each variable indicates different weights. Which weight can you find? Find it.
SOVA2 [1]

Answer:

We can only be certain that <em>a</em> weighs 12.

There are infinitely many possiblities for <em>b</em> and <em>c</em>.

Step-by-step explanation:

We have the equation:

a+b+c+a+c+b+a+c=12+a+a+b+b+c+c+c

Each variable indicates a weight.

We would like to determine the weights of each variable (if possible).

First, we can rearrange the equation to acquire:

(a+a+a)+(b+b)+(c+c+c)=12+(a+a)+(b+b)+(c+c+c)

We can combine like terms:

3a+2b+3c=12+2a+2b+3c

Notice that both sides have 2<em>b</em> and 3<em>c</em>. Therefore, it is possible for us to cancel them since each nullify the other side. So, we will subtract 2<em>b</em> and 3<em>c</em> from both sides. This yields:

3a=12+2a

Therefore, we can solve for <em>a</em>. Subtract 2<em>a</em> from both sides:

a=12

Hence, the weight of <em>a</em> is 12.

Using the newly acquired information, we can go back to our simplified equation:

3a+2b+3c=12+2a+2b+3c

Since <em>a</em> is 12:

3(12)+2b+3c=12+2(12)+2b+3c

Evaluate:

36+2b+3c=12+24+2b+3c

Simplify:

36+2b+3c=36+2b+3c

We can subtract 36 from both sides:

2b+3c=2b+3c

As you can see, this is a true statement.

Since this is a true statement, there are infinitely many possible values for <em>b</em> and <em>c</em>.

Therefore, the only weight we are <em>certain</em> of knowing is weight <em>a</em> weighing 12.

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