Bricks are stacked according to the pattern shown below. Each row contains one more brick than the row above it.
Part A
Determine a quadratic function to model the total number of bricks in the stack,
f(x), given a number of rows, m.

Part A

The Quadratic Function to model the total number of bricks the stack is f(x) = x²

Part B

The number of bricks we will have for 50 rows is 2500 bricks

Step-by-step explanation:

Part A

Let the Quadratic Function to model the total number of bricks the stack, f(x), where x is the number of rows be f(x) = a·x² + b·x + c

We have;

When x = 0, f(x) = 0, therefore, f(0) = a×0² + b×0 + c = 0 + 0 + c = 0

∴ c = 0

When x = 1, f(x) = 1, therefore, f(1) = a×1² + b×1 + 0 = 1

∴ a + b = 1...(1)

When x = 2, f(x) = 4, therefore, f(2) = a×2² + b×2  = 4

∴ 4·a + 2·b = 4...(2)

Multiply equation (1) by 2, and subtract it from equation (2) gives;

4·a + 2·b - 2×(a + b) = 4 - 2 × 1

2·a = 2

a = 1

From equation (1), we have;

b = 1 - a = 1 - 1 = 0

b = 0

∴ f(x) = a·x² + b·x + c = 1·x² + 0·x + 0 = x²

Therefore, the Quadratic Function to model the total number of bricks the stack, f(x), where x is the number of rows be f(x) = x²

Part B

If there are 50 rows in the stack, then we have, x = 50

Therefore;

f(50) = 50² = 2500

The number of bricks we will have for 50 rows is f(50) = 2500 bricks

(just found it on )

a.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?