Question

If C-t^(2)-4t-7 and D-2t+6, find an expreasion that equals 3C-3D in standard form.

Answer 1

To find 3C - 3D, multiply each expression by 3 and subtract 3D from 3C. The final expression in standard form is 3t^2 + 6t + 3.

Step-by-step explanation:

To find an expression that equals 3C - 3D in standard form, first we need to define expressions C and D. Given that C equals t2 + 4t + 7 and D equals 2t + 6, we multiply each by 3 to get 3C and 3D:

• 3C = 3(t2 + 4t + 7) = 3t2 + 12t + 21
• 3D = 3(2t + 6) = 6t + 18

Subtracting 3D from 3C gives us the following:

3C - 3D = (3t2 + 12t + 21) - (6t + 18)

Now subtract the terms:

= 3t2 + 12t - 6t + 21 - 18

= 3t2 + 6t + 3

The final expression in standard form is 3t2 + 6t + 3.

Answer 2

The expression that equals 3C - 3D in standard form is written as 3t² - 18t - 39 = 0.

The standard form of writing quadratic equations.

To determine the standard form of the quadratic equation, we need to follow the following steps. Let us find the difference in the algebraic expressions given.

Given that:

• C = t² - 4t - 7; 3C will be = 3(t² - 4t - 7)
• D = 2t + 6; 3D will be = 3(2t + 6)

3C = 3t² - 12t - 21

3(2t + 6) = 6t + 18

Now, the difference of these algebraic equation is:

= (3t² - 12t - 21) - (6t + 18)

Combine like terms

= 3t² - 12t - 6t - 21 - 18

= 3t² - 18t - 39

Since the standard form of a quadratic equation is represented as: ax² + bx + c = 0. Then the expression that equals 3C - 3D in standard form is written as 3t² - 18t - 39 = 0.