100
8x - 2y-120 = 0 subsitute the x with zero to find the y intercept 0x - 2y-120=0 -2y-120=0 -2y=120 y=60 substitue the y with zero to find the x intercept 8x-120=0 8x=120 x=120/8 x=15 lol hopefully i did this rite... sorry if i didnt XD i have a coordinate geometry test tomorrow... lol
80=4x, z=6x. find Z. x=80/4, x=20. z=6x20 z=120 answer= 120
y = 10x
X = 135 and y = 15 Solved by addition and substitution
160
We have the numbers x and y. x+y=120 x-y=50 2x=170 x=85 y=35
To find the y-intercept of the equation ( y = 12x + 120 ), set ( x = 0 ). Substituting, we get ( y = 12(0) + 120 = 120 ). Therefore, the y-intercept is ( (0, 120) ).
8x - 2y-120 = 0 subsitute the x with zero to find the y intercept 0x - 2y-120=0 -2y-120=0 -2y=120 y=60 substitue the y with zero to find the x intercept 8x-120=0 8x=120 x=120/8 x=15 lol hopefully i did this rite... sorry if i didnt XD i have a coordinate geometry test tomorrow... lol
200
If you meant 4x=3y-120 ==> 4x +120 = 3y ==> 4/3x+40=y In y=4/3x+40 4/3 is the slope (y will be increased by 4 when x is increased of 3) and +40 is the y intercept (when x is zero y is +40)
80=4x, z=6x. find Z. x=80/4, x=20. z=6x20 z=120 answer= 120
50
Since ( x ) varies directly with ( y ) and ( z ), we can express this relationship as ( x = k \cdot y \cdot z ) for some constant ( k ). Given that ( x = 120 ) when ( y = 5 ) and ( z = 8 ), we can find ( k ) as follows: [ 120 = k \cdot 5 \cdot 8 \implies k = \frac{120}{40} = 3. ] Now, to find ( x ) when ( y = 6 ) and ( z = 2 ): [ x = 3 \cdot 6 \cdot 2 = 36. ] Thus, ( x = 36 ).
To find two numbers that add to 24 and multiply to -160, we can set up a system of equations. Let's call the two numbers x and y. We have the equations x + y = 24 and x * y = -160. By solving these equations simultaneously, we can find that the two numbers are 20 and -8. This is because 20 + (-8) = 12 and 20 * (-8) = -160.
y = 10x
290
Maximum area for a rectangle of a given perimeter is a square so side would be 30m and area 900 sq m Math behind it: We have two variables X and Y (width and height) The perimeter of the rectangle is 120, therefore 2X + 2Y = 120 Now that we have an equation we can get rid of one of the variables like so: 2X + 2Y = 120 => 2Y = 120 - 2X => Y = 60 - X We want to optimize the area (A), which is X*Y: A = X * Y => Substitute the Y -> A = X * (60 - X) => A = 60X - X^2 To find the maximum area we'll have to find where the derivative A' zero. A = 60X - X^2 => Find the derivative -> A' = 60 - 2X A' = 0 => 60 - 2X = 0 => -2X = -60 => X = 30 Filling in the X to get the Y we get: Y = 60 - X => Y = 60 - 30 => Y = 30 Now that we have the X and the Y we can calculate the maximum area which is: X * Y => 30 *30 => 900