Answer:
Step-by-step explanation:
1. Approach
Since it is given that the garden box is a rectangle, then the opposite sides are congruent. One can use this to their advantage, by setting up an equation that enables them to solve for the width of the rectangle. After doing so, one will multiply the width by the given length and solve for the area.
2. Solve for the width
It is given that the garden box is a rectangle. As per its definition, opposite sides in a rectangle are congruent. The problem gives the length and the perimeter of the rectangle, therefore, one can set up an equation and solve for the width.
Substitute,
Conver the mixed number to an improper fraction. This can be done by multiplying the "number" part of the mixed number by the denominator of the fraction. Then add the result to the numerator.
Inverse operations,
3. Solve for the area
Now that one has solved for the width of the box, one must solve for the area. This can be done by multiplying the length by the width. Since the width is a fraction, one must remember, that when multiplying an integer by a fraction, one will multiply the integer by the numerator (the top of the fraction), and then simplify by reducing the fraction, if possible. Reducing the fraction is when one divides both the numerator and the denominator by the GCF (Greatest Common Factor).
Substitute,
it's A because you put your Decimals in order and round the nearest tenths.
Answer:
the lower right matrix is the third correct choice
Step-by-step explanation:
Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).
Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.
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We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.
The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...
![\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C1%2B2%261-3%261-1%261000%2829%2B1%29%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C3%26-2%260%2630000%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D)
Already, we see that the second row matches that in the lower right matrix.
The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.
The correct choices here are the two you have selected, and <em>the lower right matrix</em>.
Answer:
1. Answer is 162 2. Answer is 159.2
Step-by-step explanation:
In the first one we calculate both triangles by using and multiply it by 2 since we have 2 triangles
then we calculate the rectangle
and then we put them all together and get the answer which is 162.
in the second one we calculate the trapezoid then rhombus, then we put them together and get our answer which is 159.2
