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lesantik [10]
3 years ago
7

0(-4,1), and Y (-2, 3).

Mathematics
1 answer:
Mashutka [201]3 years ago
6 0

Answer: If you trying to find the distance d=\sqrt{16} or 4. If you trying to find midpoint its MP= (-3,2)

Step-by-step explanation:Distance formulad=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{2})^{2}  }

d=\sqrt{(-2-(-4))^{2}+(3-1})^{2}  } -2-(-4)=2     3-1=2

d=\sqrt{2^{2}+2^{2}  } 2 squared is 4

d=\sqrt{4+4}  equals d=\sqrt{16 } or 4 simplyfied

Midpoint formula MP= (\frac{x_{1}+x_{2}  }{2}), (\frac{y_{1}+y_{2}  }{2} )

MP= (\frac{-4+-2  }{2}), (\frac{1+3  }{2} ) -4+-2=-6     1+3=4

MP= (\frac{-6}{2} ), (\frac{4  }{2} ) -6/2= -3    4/2= 2

MP= (-3,2)

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5(t - 3) -2t = -30 what is the answer?
maxonik [38]

Answer:

t = -5

Step-by-step explanation:

Solve for t:

5 (t - 3) - 2 t = -30

Hint: | Distribute 5 over t - 3.

5 (t - 3) = 5 t - 15:

5 t - 15 - 2 t = -30

Hint: | Group like terms in 5 t - 2 t - 15.

Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:

(5 t - 2 t) - 15 = -30

Hint: | Combine like terms in 5 t - 2 t.

5 t - 2 t = 3 t:

3 t - 15 = -30

Hint: | Isolate terms with t to the left hand side.

Add 15 to both sides:

3 t + (15 - 15) = 15 - 30

Hint: | Look for the difference of two identical terms.

15 - 15 = 0:

3 t = 15 - 30

Hint: | Evaluate 15 - 30.

15 - 30 = -15:

3 t = -15

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 3 t = -15 by 3:

(3 t)/3 = (-15)/3

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3/3 = 1:

t = (-15)/3

Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.

The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:

Answer: t = -5

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