<h2>
Answer:
<u><em>
</em></u></h2>
Step-by-step explanation:
Step 1: Multiply the whole number part (1) by the denominator (8).
1 × 8 = 8
Step 2: Add the product from Step 1 (8) to the numerator (4).
8 + 4 = 12
Step 3: Write that result (12) above the denominator. So,
Step 4: The fraction 
Can be reduced by dividing both numerator and denominator by the GCD(12,8) = 4. Thus,
Answer: So lets say that the width is w. The length(L) is 4 meters greater than 3 times the width. 3 times the width would be 3w, and then 4 meters greater than that would be 3w+4. You have two widths and two lengths to a rectangle that must add up to equal 72. So 2w + 2L =72. But remember that one L equals 3w+4
2(w) + 2(3w+4) = 72
2w+6w+8=72
8w+8=72
8w=64
w=8
L=3w+4
L=3(8)+4
L=24+4
L=28
So the dimensions are 8 meters by 28 meters
To check 2(8)+2(28)=72
Answer:
None
Step-by-step explanation:
Answer:
The solution to the equation system given is:
- <u>x = 2</u>
- <u>y = -1</u>
Step-by-step explanation:
First, we must know the equations given:
- 2x + 3y = 1
- 3x + y = 5
Following Crammer's Rule, we have the matrix form:
![\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C3%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Now we solve using the determinants:
![x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D%7D%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C3%261%5Cend%7Barray%7D%5Cright%5D%20%7D%20%3D%5Cfrac%7B%281%2A1%29-%285%2A3%29%7D%7B%282%2A1%29-%283%2A3%29%7D%20%3D%20%5Cfrac%7B1-15%7D%7B2-9%7D%20%3D%5Cfrac%7B-14%7D%7B-7%7D%20%3D%202)
![y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D%7D%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C3%261%5Cend%7Barray%7D%5Cright%5D%20%7D%20%3D%5Cfrac%7B%282%2A5%29-%283%2A1%29%7D%7B%282%2A1%29-%283%2A3%29%7D%3D%5Cfrac%7B10-3%7D%7B2-9%7D%20%3D%5Cfrac%7B7%7D%7B-7%7D%3D-1)
Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:
- 2x + 3y = 1
- 2(2) + 3(-1)= 1
- 4 - 3 = 1
- 1 = 1
And, with the second equation:
- 3x + y = 5
- 3(2) + (-1) = 5
- 6 - 1 = 5
- 5 = 5
<span>Find all pairs of consecutive even integers whose sum is greater than 73 but less than 79.
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