The equation of a straight line is 2y = 3x + 4.

a. Find the graduent of this line.

b. Find the co-ordinates of the point where the line crosses the y-axis.

Please answer fast

The standard form for expressing the equation of a line is y = mx + b where;

Given the equation

2y = 3x + 4

Rewrite in standard format

y = 3/2 x + 4/2

y = 3/2 x + 2

Get the gradient

mx = 3/2 x

m = 3/2

Hence the gradient of the line is 3/2

b) The line crossed the y-axis when x = 0. Substitute x = 0 into the expression to get the required coordinate

y = 3/2 x + 2

y  = 3/2(0) + 2

y = 2

Hence the co-ordinates of the point where the line crosses the y-axis is at (0, 2)

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a. 3/2.

b. (0, 2).

Step-by-step explanation:

a. 2y = 3x + 4       Divide through by 2:

y = 3/2 x + 2

Comparing this to the general form y = mx + c where m = the gradient  we see  that the gradient = 3/2.

b.  When the line crosses the y-axis x = 0 so we have the equation:

2y = 3(0) + 4

2y = 4

y = 2.

So the  required point is (0, 2).

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

not sure but it may be b, as i said i'm not sure.