Full Grade 12 test

Equal tangent values repeat every 180°. Write the general inner-angle solution, rearrange for x and select roots in the interval. Check both endpoints, then add the solutions.

Find an antiderivative by integrating each term. Evaluate it at both integration bounds and subtract the lower-bound value from the upper-bound value. The integration constant cancels.

Find the domain and critical points, then the derivative signs on intervals. f′ > 0 means increasing and f′ < 0 decreasing. A change from − to + gives a minimum; + to − a maximum. f′ = 0 alone does not guarantee an extremum. For f′ = 2(x − r), the minimum is at r.

Check both requirements: the endpoints lie on the specified lines and the segment is perpendicular to both. Perpendicularity to only one line is insufficient.

Find the x-intercept and split the interval there. Calculate both region areas as positive values. A single signed integral does not always equal the total geometric area.

Square the side lengths to obtain base areas. Use V = h(S₁ + √(S₁S₂) + S₂) / 3, or reconstruct the complete and removed pyramids and subtract their volumes.

Find the domain and critical points, then the derivative signs on intervals. f′ > 0 means increasing and f′ < 0 decreasing. A change from − to + gives a minimum; + to − a maximum. f′ = 0 alone does not guarantee an extremum. For f′ = 2(x − r), the minimum is at r.

A section is the intersection of a solid with a plane. Identify its shape and dimensions before calculating. An axial section plane passes through the rotation axis; section area is not the total surface area.

Rewrite the fraction as a negative power of x, apply the power rule and differentiate to check. The logarithm rule applies to exponent −1, not to every fraction.

A cylinder axial section is a rectangle with sides 2r and h. Its area is 2rh. Use the diameter, not just the radius.

Determine the sign of the difference between x and the approach point on the specified side. Remove the absolute-value signs accordingly, cancel and finish the calculation.

Set the two model predictions equal: a₁x + b₁ = a₂x + b₂. Collect the x terms on one side, the constants on the other, and solve the resulting linear equation.

Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.

Cylinder volume is circular base area × height: V = πr²h. Use π = 3.14 and cm³.

First find the correlation coefficient: |r| = √R². Determine its sign from the direction of the regression line. Then use the slope formula a = r · sᵧ / sₓ. Keep the sign of r.

Apply the product rule. Differentiating the exponential adds the derivative of its linear exponent as a factor. Add both product-rule terms and substitute x = 0.

The correlation sign gives direction: r > 0 positive and r < 0 negative. Compare linear strength using |r|: closer to 1 means stronger. A negative r can indicate a strong relationship.

The dependent variable is the response predicted by the model. Assess fit using R², residual patterns, context and prediction range. Correlation alone does not prove causation; a third factor may explain it.