Question

The graph of linear equation 3x + 2y = 16 cuts y-axis at
(2,8)
(2, -4)
(8,0)
(0,8)​

Answer 1

Answer:

Since The graph of linear equation 3x + 2y = 16 cuts y-axis

So

x = 0

$\displaystyle \: 3 \times 0 + 2y = 16$

$2y = 16$

$y = \frac{16}{2} = 8$

So the required point is (0,8)

.

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