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3 years ago

Find inverse equation of f(x)=(2x-3)/(x+1)

Mathematics

1 answer:

Ksivusya [100]3 years ago

5 0

Answer:

To find the inverse, you still interchange the variables

f - 1 (x) = 3 + x/2 - x is the Answer Hope this helps :)

Step-by-step explanation:

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A chocolate chip cookie recipe that makes 24 cookies uses 3/4 cups of sugar. if raphael wants to make 48 cookies, how much sugar

ladessa [460]

1.5 cups

Step-by-step explanation:

3x2=6

4 goes into 6 once and leaves 2 which is half of 4

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3 years ago

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Find x from given equations sin30 + 2 cot^2 30 + x cos^2 30= 8+ tan^2 45 + cos 60

julia-pushkina [17]

Answer

1/2+2×(sqrt3)^2+x.(sqrt3/2)^2=8+1^2+1/2

1/2+2×3+x.3/4=8+1+1/2

1/2-1/2+6+x.3/4=9

x.3/4=9-6

x.3×4=3

x.3=3×4

x.3=12

x=12/3

x=4.

so, the value of x is 4.

hope it helps you.

4 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

a circle is divided into 18 equal parts how many degrees is the angle for each part? how many degrees is the angle for 5 parts?

Ket [755]

There are 360 degrees in a circle, and we have 18 pieces, so we need to see how many times 18 goes into 360. We can find this out by dividing.

360/18=20

You can check this by multiplying 20 and 18 (it equals 360)!

So, each fraction of the circle will be 20 degrees.

If you take 5 of these 20 degree pieces, you'll need to multiply them by 20 to see how many degrees they'd be.

You need to multiply by 20 because each piece is 20 degrees, and we need to find how many degrees 5 pieces is. It's the same as doing
20+20+20+20+20! :)

20*5=100. 5 parts will be 100 degrees.

Hope I helped! :)

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3 years ago

What should you multiply the top equation by so that y is eliminated when the

VLD [36.1K]

Your answer is D. 2 :)

8 0

3 years ago

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