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3 years ago

Find the midpoint of the line on the coordinate plane with the given endpoints below.

Mathematics

2 answers:

tigry1 [53]3 years ago

4 0

Answer:

Then the midpoint is (2, -2)

Step-by-step explanation:

We treat the x- and y-coordinates separately, first finding the average (mean) of the x-coordinates and then the average of the y-coordinates:

xm = (-1 + 5)/2 = 2

ym = (1 - 5)/2 = -2

Then the midpoint is (2, -2)

Marianna [84]3 years ago

3 0

Answer:

(2,-2)

Step-by-step explanation:

Two points are given to us and we need to find the midpoint of those two points. The points are

(-1,1) and (5,-5) . Let us say we have two points ( x1,y1) and ( x2 , y2) . The midpoint will be ,

$\implies Midpoint =\bigg( \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \bigg) \\\\\implies Midpoint = \bigg( \dfrac{-1+5}{2},\dfrac{1-5}{2}\bigg)\\\\\implies Midpoint = \bigg( \dfrac{4}{2},\dfrac{-4}{2}\bigg)\\\\\implies \red{Midpoint = ( 2 , -2) }$

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Serga [27]

Answer:

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Step-by-step explanation:

1) A segment is a line portion with two end points

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3) Collinear points are points on the same line

3) A quadrilateral is a four sided figure while rectangle has 90° interior angles

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2 years ago

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3 years ago

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Answer:

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2 years ago

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Step-by-step explanation:

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3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

3 years ago