For 5th grade which equation provides the best estimate of the product of 1.7 and 3.2
1 answer:
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<h3>Answer: k = 7</h3>
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Explanation:
There are probably a number of ways to approach this problem, but row reduction was the only method I could think of at the moment.
If you were to follow the steps shown in the attached image, then you'll be on the process of applying row reduction. The last step in that diagram isn't in full REF (row echelon form), but that doesn't technically matter.
What does matter is the 2k-14 entry in the bottom row. If that entry was 0, while the entry just to the right of it was nonzero, then this would lead the entire system of equations to be inconsistent. This is because the bottom equation would be in the form 0x+0y+0z = m, where m is some nonzero constant. As you can see, that equation would simplify to 0 = m; however, m is nonzero, so we have a contradiction.
If 2k - 14 were 0, then
2k - 14 = 0
2k = 14
k = 14/2
k = 7
This is the only k value in which the system is inconsistent. In other words, the system wouldn't have any solutions with this k value. You can verify this through completing the row reduction with k = 7 (it should be far easier now that we can nail down a fixed k value) and find that you'll get a contradiction. You could also use substitution to find an inconsistency would arise when k = 7.
Side notes:
- You don't need to do full RREF, though you can if you want. REF should be sufficient.
- I drew each matrix as a grid of boxes to help separate the terms. This is also done to space out each step. Usually those grid lines aren't present.
Answer:
Formula of Trapezoid:
A = (a + b) × h / 2
The formula can be derived in different ways. for now, we have discussed two ways:
1. By using the formula of a triangle
2. By dividing into different sections
Step-by-step explanation:
1. By using the formula of a triangle
One of the ways to explain a formula for an area of a trapezoid using a formula for a triangle can be as follows.
Assume a trapezoid PQRS with lower base SR and upper base PQ (they are parallel) and sides PS and QR.
The image is attached below.
Connect vertices P and R with a diagonal.
Consider triangle ΔPQR as having a base PQ and an altitude from vertex R down to point M on base PQ (RM⊥PQ).
Its area is
S1=
Consider triangle ΔPRS as having a base SR and an altitude from vertex P up to point N on-base SR (PN⊥SR).
Its area is
S2=
Altitudes RM and PN are equal and constitute the distance between two parallel bases PQ and SR.
They both are equal to the altitude of the trapezoid h.
Therefore, we can represent areas of our two triangles as
S1=
S2=
Adding them together, we get the area of the whole trapezoid:
S=S1+S2=
which is usually represented in words as "half-sum of the bases times the altitude".
2. By dividing into different sections
Trapezoid PQRS is shown below, with PQ parallel to RS.
Figure 1 - Trapezoid PQRS with PQ parallel to RS(image is attached below.)
We are going to derive the area of a trapezoid by dividing it into different sections.
If we drop another line from Q, then we will have two altitudes namely PT and QU.
Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle. (image is attached below.)
From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that
=>
Simplifying, we have
=>
Factoring we have,
=>
But, is equal to
, the longer base of our trapezoid.
Hence,
We have discussed two ways by which we can derive area of a trapezoid.
Read to know more about Trapezoid
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