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Find y 160 120 and 50?

Updated: 4/28/2022
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Q: Find y 160 120 and 50?
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What are 2 numbers whose sum is 120 and whose difference is 50?

We have the numbers x and y. x+y=120 x-y=50 2x=170 x=85 y=35


Find x-intercept and y-intercept of the line given the equation 8x-2y-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


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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


2 numbers can be added to equal 7 and multiplied to equal -120?

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