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2 years ago

Help me out Whts 4+2y=16

Mathematics

2 answers:

FinnZ [79.3K]2 years ago

8 0

Answer: y=6

Step-by-step explanation:

2y=16-4

2y=12

y=12/2

y=6

Triss [41]2 years ago

7 0

Answer:

y=6

Step-by-step explanation:

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Jaz was 43 inches tall. Eighteen months later, she was 52 inches tall. The constant rate of change for Jaz's height was .

maxonik [38]

Answer:

Constant rate of change for Jaz height will be 0.5 inch per month

Step-by-step explanation:

Jaz present height is 43 inches.

Eighteen months later her height will be 52 inches.

Now we know rate of change will be represented by the formula

Rate of change of height =

= 0.5 inch per month

The constant rate of change for Jaz height will be 0.5 inch per month.

Read more on Brainly.com - brainly.com/question/2012372#readmore

7 0

3 years ago

3 5 y = mx + c Find the value of y when m= -2, r = -7 and c= -3.

mario62 [17]

Answer:

This assume the equation is 35y = mx + c, and that r is meant to be x.

Step-by-step explanation:

35y = mx + c

35y = -2(-7) - 3

35y = 14 - 3

35y = 11

y = (11/35)

Please check the formatting of your question.

6 0

2 years ago

Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph

kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0 and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of f(y) versus y

Looking at the given differential equation

$\frac{dy}{dt} = e^{y} - 1$ for -∞ < $y_{o}$ < ∞

We can say let $\frac{dy}{dt}$ = f(y) = $e^{y} - 1$

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

To fin the critical point we have to set $\frac{dy}{dt}$ = 0

This means $e^{y} - 1$ = 0

For this to be possible $e^{y}$ = 1

which means that $e^{y}$ = $e^{0}$

which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.

In each of the intervals bounded by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)