All categories

3 years ago

Given x(y + 2) = 3x + y, express y in terms of x.

Mathematics

1 answer:

vodka [1.7K]3 years ago

6 0

Answer:

y = $\frac{x}{x-1}$

Step-by-step explanation:

Given

x(y + 2) = 3x + y ← distribute parenthesis on left side

xy + 2x = 3x + y ( subtract 2x from both sides )

xy = x + y ( subtract y from both sides )

xy - y = x ← factor out y from each term on the left side )

y(x - 1) = x ← divide both sides by (x - 1)

y = $\frac{x}{x-1}$

You might be interested in

X/w = z/y². solve for y

jek_recluse [69]

Answer:

Step-by-step explanation:

We can simplify (cross multiply) to get $xy^2=wz$

Then we easily simplify to get $y=\sqrt{\frac{wz}{x}},\:y=-\sqrt{\frac{wz}{x}};\quad \:w\ne \:0,\:x\ne \:0$

8 0

3 years ago

La señora esta caminando en la calle y saludo a todos

grandymaker [24]

Hola, necesita ayuda con translation?
La señora esta caminando en la calle y Saludo a todos.
The woman is walking in the street and greets everyone.

5 0

3 years ago

What is 3/7 in words

Marrrta [24]

Three sevenths !!!!!!

4 0

3 years ago

Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour

UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

$C(t) = C(0)e^{rt}$

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So $C(0) = 110$

Husband

Half life of 4 hours. So

$C(4) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{4r}$

$e^{4r} = 0.5$

Applying ln to both sides

$\ln{e^{4r}} = \ln{0.5}$

$4r = \ln{0.5}$

$r = \frac{\ln{0.5}}{4}$

$r = -0.1733$

So for the husband

$C(t) = 110e^{-0.1733t}$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.1733t}$

$C(11) = 110e^{-0.1733*11} = 16.35$

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

$C(10) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{10}$

$e^{10r} = 0.5$

Applying ln to both sides

$\ln{e^{10r}} = \ln{0.5}$

$10r = \ln{0.5}$

$r = \frac{\ln{0.5}}{10}$

$r = -0.0693$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.0693t}$

$C(11) = 110e^{-0.0693*11} = 51.33$

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0

3 years ago

6w = 21<br><br> Solve for w<br><br> Please show steps

brilliants [131]

Answer:

w=21/6

w=3.5

Step-by-step explanation:

I hope it's helps you

4 0

3 years ago

Given x(y + 2) = 3x + y, express y in terms of x.​

Given x(y + 2) = 3x + y, express y in terms of x.