Answer:
<h2>
8 units²</h2>
Step-by-step explanation:
Let side of the square = a
The , Area of square = a²
Now, Midpoint of Diagonal DB is E
And DE = 2 units
So, DB = 2 DE = 2 × 2 = 4 units
Now, using Pythagoras theorem in ∆ BCD
DB² = DC² + BC²
plug the values
Collect like terms
Evaluate the power
Swipe the side of the equation
Divide both sides of the equation by 2
Calculate
Therefore, The area of the square is 8 sq.units.
Hope this helps..
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4x - 5y = -12
-x + 5y = 18
Add the equations together:
4x + -x = 3x
-5y + 5y = 0
-12 + 18 = 6
So now you have:
3x = 6
Divide both sides by 3:
X = 2
Now you have the value of x, replace x with 2 in one of the equations and solve for y:
4(2) -5y = -12
8 -5y = -12
Subtract 8 from both sides:
-5y = -20
Divide both sides by -5:
Y = 4
The solution is x = 2 and y = 4 which is written as (2,4)
Multiply each hourly rate by x ( time to complete the work) add it to the service call for each and then set the equations equal:
50 + 40x = 30 +45x
Subtract 30 from both sides:
20 + 40x = 45x
Subtract 40x from both sides:
20 = 5x
Divide both sides by 5
x = 4
The length is 4 hours.
You do 4 x 56 = 224. It’s just multiplication.
When you calculate 2x-3=2x-5 it’s going to lead as -3=-5, therefore this statement is inaccurate and doesn’t have a value of x, it doesn’t even have a solution.
