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svetoff [14.1K]
4 years ago
14

3(2c+d)-4(c-d)+d squared when c=1 d=3 please explain steps

Mathematics
1 answer:
steposvetlana [31]4 years ago
4 0

Answer:

3(2c+d)-4(c-d)+d^2.

6c+3d-(4c+4d)+d^2.

6×1+3×3-4×1+4×3+9.

6+9-4+12+9=32.

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Answer:

1

Step-by-step explanation:

Choose two points on the line:  (-5, 0) and (0, 5)

Let the first point (x_1,y_1) = (-5, 0)

Let the second point (x_2,y_2) = (0, 5)

Use the gradient formula:

\textsf{gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{5-0}{0-(-5)}=1

----------------------------------------------------------------------------------------------------

If you want to find the equation of the line, use the point-gradient form of a linear equation:  y-y_1=m(x-x_1)

(where m is the gradient and (x_1,y_1) is a point on the line)

Substituting the values into the formula:

\implies y-0=1(x-(-5))

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\implies y=x+5

Therefore, the equation of the line is y = x + 5

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2 years ago
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Here are the answer to your question

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