Answer:
y=4
Step-by-step explanation:
Let's solve your system by substitution.
x+y=7;x+2y=11
Step: Solve x+y=7 for x:
x+y=7
x+y+−y=7+−y(Add -y to both sides)
x=−y+7
Step: Substitute−y+7forxinx+2y=11:
x+2y=11
−y+7+2y=11
y+7=11(Simplify both sides of the equation)
y+7+−7=11+−7(Add -7 to both sides)
y=4
Step: Substitute 4 for y in x=−y+7:
x=−y+7
x=−4+7
x=3(Simplify both sides of the equation)
Answer:
x=3 and y=4
The solution for the problem is:
I will first get the first five terms so that I could easily locate the third term of this problem:So, substituting the values:
T(1) = 1^2 = 1T(2) = 2^2 = 4T(3) = 3^2 = 9T(4) = 4^2 = 16T(5) = 5^2 =25
So the third terms is T(3) = 3^2 = 9
This is the answer Hope it helps :)
14. [-27, 0, 1.728, 125]
15. [ 18, 15, 13.8, 10]
16. [-3.9, 0, 1.56, 6.5]
17. [9, 0, 1.44, 25]
18. [12, 0, -4.8, -20]
19. [10, 1, 2.44, 26]
20. [64, 25, 14.44. 0]
21. [-22, -4, 3.2, 26]
That's a trapezoid,
![A = \frac 1 2 (b+B)h](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%201%202%20%28b%2BB%29h)
![A = \frac 1 2 (4+12) 4 = 32](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%201%202%20%284%2B12%29%204%20%3D%2032)
Answer: 32
But say we didn't know the formula. We can chop out the 4 by 4 square and we have two 45/45/90 triangles side 4 left over, so two 4 by 4 squares,
A = 2(4)(4) = 32
We can think of this as 12 by 4 rectangle with two 4 by 4 right triangles chopped out:
A = 12(4) - 2(1/2)(4)(4) = 48 - 16 = 32