What point divides the line segment between (10,-2) and (-6,6) into a ratio of 5:3

Mathematics

1 answer:

olganol [36]2 years ago

3 0

Answer: (0,3)

Step-by-step explanation:

You follow the formulas to find the x and y of the dividing points.

Xp= x1+ a/a+b (x2-x1)

Xp= 10+5/5+3(-6-10)

Xp=10+5/8(-16)

Solve the problem above and you end up with “0” as your x.

Yp= y1+ a/a+b (y2-y1)

Yp= -2+ 5/8 (6+2)

Yp= -2+ 5/8 (8)

Solve this problem above and you end up with “3” as your y. Therefore, the point that divides the line segment between (10, -2) and (-6,6) into a ratio of 5:3 is (0, 3).

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3.25x-1.5y=1.25 13x-6y=10

Kaylis [27]

Step-by-step explanation:

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '1.5y' to each side of the equation.

3.25x + -1.5y + 1.5y = 1.25 + 1.5y

Combine like terms: -1.5y + 1.5y = 0.0

3.25x + 0.0 = 1.25 + 1.5y

3.25x = 1.25 + 1.5y

13x + -6y = 10

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '6y' to each side of the equation.

13x + -6y + 6y = 10 + 6y

Combine like terms: -6y + 6y = 0

13x + 0 = 10 + 6y

13x = 10 + 6y

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4 0

2 years ago

Read 2 more answers

Evaluate the line integral 2 + x2y ds where c is the upper half of the circle x2 + y2 = 1.

goldfiish [28.3K]

Parameterize C by

r(t) = 〈x(t), y(t)〉 = 〈cos(t), sin(t)〉

with 0 ≤ t ≤ π. Then the line integral is

$\displaystyle \int_C (2+x^2y)\,\mathrm ds = \int_0^\pi (2+\cos^2(t)\sin(t))\left\|\mathbf r'(t)\right\|\,\mathrm dt \\\\ = \int_0^\pi (2+\cos^2(t)\sin(t)) \,\mathrm dt = \boxed{\frac23+2\pi}$