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suter [353]
3 years ago
9

Given A(-3, 6) and B(0, 6), what are the coordinates of point P that lies on segment AB, such that AP:PB = 2:1? A: (2, -2) B: (-

2, 0) C: (-1, 6) D: (-1, 3)
Mathematics
1 answer:
DerKrebs [107]3 years ago
7 0

Answer:

(-1 1/2, 6)

Step-by-step explanation:

The mid point between A(-3,6) and B(0,6) represents the coordinates of point P, which is (x,y)

Hence,the coordinates, that is, AP: x = (-3 + 0)/2 = -3/2 = -1 1/2

also, BP: y = (6 + 6)/2 = 6

Therefore, the coordinates of point P = ( -1 1/2, 6)  

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How do u subtract (3x+5y-4)_(4×+11)
AleksAgata [21]
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4 years ago
H(x)=-x^2+6x. So what is the value of h(2)
crimeas [40]

✩ Answer:

 ✧・゚: *✧・゚:*✧・゚: *✧・゚:*✧・゚: *✧・゚:*

  \bold{Hello!}\\\bold{Your~answer~is~below!}

✩ Step-by-step explanation:

 ✧・゚: *✧・゚:*✧・゚: *✧・゚:*✧・゚: *✧・゚:*

✺ Quadratic polynomials can be factored using the transformation ax^2+bx+c=a(x-x_{1})(x-x_{2} ), where x_{1} and x_{2} are the solutions of the quadratic equation ax^2+bx+c=0:

  • -x^2+6x=0

✺ All equations of the form ax^2+bx+c=0 can be solved using the quadratic formula:

  • -b=\frac{+}\\\sqrt{b^2-4ac}\\~~~~~~~~~~~~~2a

✺ The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction:

  • x=\frac{\sqrt{-6\frac{+}\\\sqrt{6^2}}}{2(-1)}

✺ Take the square root of 6^2:

  • x=\frac{\sqrt{-6\frac{+}\\{6}}}{2(-1)}

✺ Multiply 2 times -1:

  • x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}

✺ Now solve the equation x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2} when ± is plus. Add -6 to 6:

  • x=\frac{0}{-2}

✺ Divide 0 by -2:

  • x=0

       -OR-

✺ Now solve the equation x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2} when ± is minus. Subtract 6 from -6:

  • x=\frac{-12}{-2}

✺ Divide -12 by -2:

  • x=6

✺  Optional : Factor the original expression using ax^2+bx+c=a(x-x_{1})(x-x_{2} ). Substitute 0 for x_{1} and 6 for x_{2}:

  • -x^2+6x=-x(x-6)

✩ Answer:

✺ <u>Factored Form</u>: x(x-6)

✺ <u>Exact Form</u>: x=6

✺ <u>Graph Point Form</u>: x=(6,0)

Hope~this~helps~and,\\Best~of~luck!\\\\~~~~~-TotallyNotTrillex

             "ටᆼට"

8 0
3 years ago
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