There’s nothing so let’s just go with C
I'm not 100% sure what you're asking, but i think it wants:
50,000 + 25,000 = 75,000 - so B is the answer
to do it you basically just round the number to the nearest 1000 and add
To find the area of the triangle, you use the formula A=bh/2. This gives you the area of 49in^2 for the triangle, eliminating choices C and D. The area of the entire square is 14^2, or 196in^2, which gives us 4 when divided by the area of the triangle, 49in^2. Therefore, the answer is B, or that the area of the triangle is 49in^2 so she can create only 4 triangles.
Answer:
Step-by-step explanation:
Solve the following equation:
![\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%5C%5C5%265%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%5C%5Cx_%7B2%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C-3%5C%5C%5Cend%7Barray%7D%5Cright%5D)
-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:
-First equation:
-Second equation:
-Choose one of the two following equations, which I choose the first one, then you solve for 
by isolating
-Subtract 
to both sides:
-Subtract 
to both sides:
-Divide both sides by 
:
-Multiply 
by 
:
-Substitute 
for in the second equation, which is :
Multiply by 
:
-Combine like terms:
-Subtract 
to both sides:
-Multiply both sides by 
:
-After you have the value of 
, substitute for 
onto this equation, which is :
-Multiply 
and 
:
-Since both and 
have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:
