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3 years ago

Find the midpoint of the segment with the following endpoints. (-8,-6) and (2, -10)

Mathematics

1 answer:

ehidna [41]3 years ago

8 0

Answer:

Step-by-step explanation:

x-coordinate of midpoint: is -8 + (2 - [-8]), or -8 + 10/2, or - 8+ 5, or x = -3.

y-coordinate of midpoint: notice that the distance from -6 to -10 is 4; the y-coordinate of the midpoint is found by starting with -6 and adding 4/2: -4

The midpoint is (-3, -4)

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Fittoniya [83]

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2 years ago

Read 2 more answers

Select all values of x that make the inequality -x + 8 ≥11 true.

mash [69]

-x + 8 ≥ 11

-x ≥ 3

x ≤ -3

Hope this helps! ;)

8 0

3 years ago

Which of the following is a linear equation x+y-2 2x+5=11 15

skelet666 [1.2K]

Answer:

x + y = 2

y = 2 - x

so it is a linear equation.

2x + 5 = 11

2x = 11 - 5

2x = 6

x = 3

it is not a function because for just one x we have many y

5 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

FIND THE AREA OF THE FIGURE BELOW

Romashka [77]

Answer:

220 squared units

Step-by-step explanation:

first, find the area of the rectangle by multiplying 8 by 20. Then, fin the area of the triangle by multiplying 6 by 20 and then dividing the result by 2.

8*20=160

6*20=120

120/2=60

Finally, add them together and you get 220 units squared.

7 0

1 year ago