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liberstina [14]

3 years ago

8

Can someone please help me with this problem

1 answer:

LenKa [72]3 years ago

5 0

Answer:

the piont was 10.15 away from the that is dead see yea

Step-by-step explanation:

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5x + 4x +1 need answer for this can anyone help?

nalin [4]

9x+1 is the answer.
I hope this helped

8 0

2 years ago

Read 2 more answers

What are the common multiples of 36 and 48

Vaselesa [24]

For 36 3 6 13. For 48 4 8 24. I think there are more but I can't think of them

7 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

Can you help me with this question?

Rufina [12.5K]

Answer:

16.2

Step-by-step explanation:

The angle internal to the triangle at B is the supplement of the one shown, so is 65°. That is equal to the angle internal to the triangle at D. Since the vertical angles at C are congruent, the two triangles are similar by the AA theorem.

Corresponding sides of similar triangles are proportional, so we can write the proportion shown in the attachment:

BC/FC = DC/AC

BC = FC(DC/AC) = 21.6(7.2/9.6)

BC = 16.2 . . . . matches the first choice

4 0

3 years ago

the remains of an ancient ball court include a rectangular playing alley with a perimeter of about 48 M. the length of the alley

Firdavs [7]

Answer:

Length is 20 m and width is 4 m.

Step-by-step explanation:

Given:

Perimeter of the rectangular alley, $P=48\textrm{ m}$

Length is 5 times the width.

Let width be $x$ .

So, as per question,

Length, $l = 5x$

Now, perimeter of rectangle is given as:

$P=2(l+b)$

Plug in 48 for $P$ , $5x$ for $l$ and $x$ for b.

$48=2(5x+x)\\48=2(6x)\\48=12x\\x=\frac{48}{12}=4$

Therefore, width is 4 m.

Length is $5x=5\times 4=20$ m.

8 0

3 years ago