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3 years ago

Determine the midpoint of the line segment whose endpoints are (12, 8) and (−8, −6).

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

Answer:

(2, 1)

Step-by-step explanation:

Given points

(12, 8) and (−8, −6)

Midpoint is

((12 - 8)/2, (8 - 6)/2) =
(4/2, 2/2) =
(2, 1)

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A winery has a vat with two pipes leading to it. the inlet pipe can fill the vat in 44 ?hours, while the outlet pipe can empty i

lapo4ka [179]

Rate of fill-up = 1/44 = 1/44 * 77/77 = 77/3388

Rate of emptying = 1/77 = 1/77 * 44/44 =44/3388

Combined Rate of Fill-up = 77/3388 - 44/3388 = 33/3388 = 3/308

Time for combined fill-up = 308/3 = 102.6667 hours = 102 hours and 40 minutes

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3 years ago

What is 8^2 X8^3 as one base ?

sergiy2304 [10]

Answer:

8^5

Step-by-step explanation:

6 0

3 years ago

Quadrilateral ABCD is given, the interior angle at vertex B is 97 degrees and the exterior angle at vertex A is 88 degrees. Line

Citrus2011 [14]

Answer:

angle 1 = 83°

angle 2 = 88°

Step-by-step explanation:

angle 1 and angle on vertex B are

pair of co interior angles ( whose sum = 180° )

so, angle 1 + angle B = 180°

angle 1 = 180° - angle B

angle 1 = 180° - 97° = 83°

hence, angle 1 = 83°

but, angle 2 and exterior of vertex A forms alternate interior angle pair ,

so , angle 2 = angle A = 88°

i hope you got it ....

5 0

3 years ago

Solve the equation using inverse operations. 5 - 2x2^2 = -15

AURORKA [14]

Answer:

x= 2 1/2

Step-by-step explanation:

Im not too sharp at math but hope this helps!

7 0

3 years ago

Read 2 more answers

ITS TIMED PLEASE HELP

Lubov Fominskaja [6]

Answer:

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

$f(x)=\frac{1}{2}x^{2}-4x+5$

Convert to vertex form

$f(x)-5=\frac{1}{2}x^{2}-4x$

Factor the leading coefficient

$f(x)-5=\frac{1}{2}(x^{2}-8x)$

$f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x-4)^{2}$

$f(x)=\frac{1}{2}(x-4)^{2}-3$

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

5 0

4 years ago