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3 years ago

Please help fast

Mathematics

1 answer:

Vladimir79 [104]3 years ago

3 0

Answer:

B: II, IV, I, III

Step-by-step explanation:

We believe the proof statement — reason pairs need to be ordered as shown below

Point F is a midpoint of Line segment AB Point E is a midpoint of Line segment AC — given

Draw Line segment BE Draw Line segment FC — by Construction

Point G is the point of intersection between Line segment BE and Line segment FC — Intersecting Lines Postulate

Draw Line segment AG — by Construction

Point D is the point of intersection between Line segment AG and Line segment BC — Intersecting Lines Postulate

Point H lies on Line segment AG such that Line segment AG ≅ Line segment GH — by Construction

II Line segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC — Midsegment Theorem

IV Line segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC — Substitution

I BGCH is a parallelogram — Properties of a Parallelogram (opposite sides are parallel)

III Line segment BD ≅ Line segment DC — Properties of a Parallelogram (diagonals bisect each other)

Line segment AD is a median Definition of a Median

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Write the number positioned at point D.

inysia [295]

Answer:

-4 3/4

Step by step explanation:

point D is 4 numbers to the left of the 0, meaning it is a negative 4. Then there is also a 3/4 of space between the 4 and point D, meaning point D is -4 3/4 away from 0.

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2 years ago

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Solve 5.4p+13.1=−2.6p+3.5 . Check your solution. p= __

maxonik [38]

5.4p +13.1 = −2.6p+3.5 .

5.4p + 2.6p = 3.5 -13.1

8.0p = -9.6

p = -9.6 / 8

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2 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

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3 years ago

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alexandr402 [8]

Answer:

The answer is 14

Step-by-step explanation:

e=2

Substitute 2 in for e

8 + 3(2)

Solve

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uysha [10]

1. true.
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3. false.
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2 years ago