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8_murik_8 [283]

3 years ago

7

P(x)=6/7x+4 What is P(42)

1 answer:

Natali5045456 [20]3 years ago

4 0

40

P(42) = (6/7)(42) + 4 = 36 + 4 = 40

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What's 1 + 1<br><br> What's 1 + 1

miskamm [114]

Answer:

the awnser would be 2

Step-by-step explantion

u have 1 apple and get 1 from a friend and that adds up to 2

4 0

3 years ago

Read 2 more answers

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

A swimming pool had 2.5 liters of water in it. Some water evaporated, and then the pool only had 2 million liters of water in it

djyliett [7]

Hi!

If I have done this correctly, then the answer is:

2 million is 80% of 2.5

That means that 20% of the water evaported.

5 0

3 years ago

What is the equation, in slope-intercept form, of the line parallel to y = 5x+2 that passes through the point (-2, 1)?

o-na [289]

Y=5x+11 using the point-slope formula and simplifying

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3 years ago

Trapezoid CDEF is congruent to trapezoid TORS. If CF and DE are parallel, which pair of sides in trapezoid TQRS MUST also be par

kondaur [170]

Answer:

D

Step-by-step explanation:

TS and QR Is your answer

6 0

3 years ago