The perimeter of the given rectangle will be ( 12x + 10 ).
<h3>What is the perimeter?</h3>
The perimeter is defined as the sum of all the sides of the given shape. For a triangle, the perimeter will be the sum of all the sides of the rectangle.
It is given that the two sides of the rectangle are (2x + 2) and (4x + 3). Then the perimeter of the rectangle will be calculated as below:-
Perimeter = Sum of all the sides of the rectangle
Perimeter = 2 ( 2x + 2 + 4x + 3 )
Perimeter = 4x + 4 + 8x + 6
Perimeter = 12x + 10
Therefore, the perimeter of the given rectangle will be ( 12x + 10 ).
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Answer: I think it’s 4
Step-by-step explanation: count the squares surrounding the triangle, 4 times 2 is eight 1/2 of that is 4
Answer:
Step-by-step explanation:
$ 8 admission and $ 2 a ride
y = 2x + 8.....where y is the total cost and x is the # of rides
table :
x = 1 , 2 , 3 , 4 ,5 , 6
y = 10, 12 , 14, 16 , 18 , 20
Answer:
The answer is "![\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-2%5C%5C%20-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B7%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%7D)
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)
In equation (i) multiply by 3 and subtract by equation (iii):
put the value of c in equation (i):
In equation (ii) multiply by 3 then subtract by equation (iv):
put the value of d in equation (iv):
The final answer is "".
