Find the coordinates of the midpoint of the segment with the given endpoints.

1 answer:

8 0

$(\frac{x_{1} + y_{1}}{2}, \frac{x_{2} + y_{2}}{2}) \\(\frac{6 + 10}{2}, \frac{-3 + 5}{2}) \\(\frac{16}{2}, \frac{2}{2}) \\(8, 1)$

You might be interested in

Find the variance of: 6 6 7 8 10 10

marta [7]

Step-by-step explanation:

I hope this helps!!!!!

7 0

2 years ago

Thabo invests R50000 at the rate of 5% per annum compound interest periodically every six months.How many years does it take to

lina2011 [118]

Answer:

50000-5%=49500 pls like

7 0

3 years ago

If H= 2f over m+1, solve the equation for f

Viefleur [7K]

H = 2f / (m+1)
[multiply by (m+1)]
h(m+1) = 2f
[divide by 2]
f = (h (m + 1)) / 2

3b / (b+2) = 12 / (b+2)
[multiply by (b+2)]
3b = 12
[divide by 3]
b = 4

3 / (6x + 1) / 2 = 8 / (x + 4) / 3
[multiply both denominators to mike one denominator]
3 / 8(6x+1) = 8 / 3(x+4)
[expand brackets]
3 / (48x + 8) = 8 / (3x + 12)
[multiply by (48x + 8)]
3 = 8(48x + 8) / (3x+12)
[multiply by (3x+12)]
3(3x +12) = 4(48x + 8)
[simplify]
9x + 36 = 192x + 32
173x = 4
x = 4 / 173

8 0

3 years ago

A tree casts a shadow x = 100 ft long when a vertical rod 6.0 ft high casts a shadow 4.0 ft long. (See the illustration.) How ta

rewona [7]

I'm not really sure If I would be much help but I'll try best k? (cries a little) So you do 100
+ 6.0
+ 4.0
_________________
11 .0

So your answer is 11.0 for the tall tree.

8 0

3 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

3 years ago