All categories

3 years ago

Can someone help explain how to work this out.

Mathematics

1 answer:

ser-zykov [4K]3 years ago

3 0

Z+2.5=5.2-3.2z
z+3.2z=5.2-2.5
4.2z=2.7
z=9/14
That's your answer and that's how you work out.

You might be interested in

Pleaseee helppp no wrong answer please answer all

yaroslaw [1]

Step-by-step explanation:

it is the right answer you have to learn inverse function . if f:A to B and is a function then new function from B to A is said to be inverse function in which each element of set a maps unique element of b

4 0

1 year ago

Whats a square + b square I don't know thats why I am asking u dude

murzikaleks [220]

Answer:

.............................

8 0

2 years ago

32 ÷ 4 x (16 x 1/2) -2

Anika [276]

32 ÷ 4 x (16 x 1/2) -2

= 32 ÷ 4 × 8 -2

= 8 × 8 - 2 = 64-2

= 62

3 0

2 years ago

Read 2 more answers

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

2 years ago

Can someone help me with this please !!

pogonyaev

add all amd divide by the 40 ???let me know if im wrong

7 0

2 years ago

Read 2 more answers