The given equations are y = 3x^2 - kx and y = 4 - 2x.

Since the point of intersection is (n, 0), we can substitute y = 0 in both equations and solve for x.

Setting y = 0 in the first equation gives us 0 = 3x^2 - kx. This simplifies to kx = 3x^2.

Setting y = 0 in the second equation gives us 0 = 4 - 2x. Solving for x gives us x = 2.

Substituting x = 2 into the equation kx = 3x^2 gives us k*2 = 3*2^2.

Solving for k gives us k = 3*2 = 6.

Therefore, the value of k is 6.

Question

The given equations are y = 3x^2 - kx and y = 4 - 2x.

Since the point of intersection is (n, 0), we can substitute y = 0 in both equations and solve for x.

Setting y = 0 in the first equation gives us 0 = 3x^2 - kx. This simplifies to kx = 3x^2.

Setting y = 0 in the second equation gives us 0 = 4 - 2x. Solving for x gives us x = 2.

Substituting x = 2 into the equation kx = 3x^2 gives us k*2 = 3*2^2.

Solving for k gives us k = 3*2 = 6.

Therefore, the value of k is 6.

Knowee AI · Accepted Answer

The given equations are y = 3x^2 - kx and y = 4 - 2x.

Since the point of intersection is (n, 0), we can substitute y = 0 in both equations and solve for x.

Setting y = 0 in the first equation gives us 0 = 3x^2 - kx. This simplifies to kx = 3x^2.

Setting y = 0 in the second equation gives us 0 = 4 - 2x. Solving for x gives us x = 2.

Substituting x = 2 into the equation kx = 3x^2 gives us k*2 = 3*2^2.

Solving for k gives us k = 3*2 = 6.

Therefore, the value of k is 6.

y=3x 2 −kxy=4−2x In the equation above, 𝑘 is a constant. One point at which the graphs of the equations intersect in the 𝑥𝑦-plane is (𝑛, 0), where 𝑛 is a constant. What is the value of 𝑘 ?