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Vsevolod [243]
2 years ago
5

Please help! I'm very confused.​

Mathematics
2 answers:
inessss [21]2 years ago
5 0

Answer:

Angle 1

Step-by-step explanation:

This is because they form the letter F.

REMEMBER CORRESPONDING ANGLES ARE EQUAL.

HOPE THIS HELPED

Murrr4er [49]2 years ago
4 0

Answer:

do not know

Step-by-step explanation:

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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)

5 0
3 years ago
What is x-2y=1, solve for y
GREYUIT [131]
The answer is X=1+2y

6 0
3 years ago
Ms. Velez will use both x gray bricks and y red bricks to build a wall around her garden. Gray bricks cost $0.45 each and red br
Alekssandra [29.7K]

It is given in the question that

Ms. Velez will use both x gray bricks and y red bricks to build a wall around her garden. Gray bricks cost $0.45 each and red bricks cost $0.58 each. She can spend up to $200 on her project, and wants the number of red bricks to be less than half the number of gray bricks.

Maximum she can spend is $200. That is

0.45x + 0.58y \leq 200&#10;\\  y< \frac{1}{2} x

And

x\geq 0 , y \geq 0

And that's the required inequalities .

7 0
3 years ago
Trapezia ABCD and PQRS are similar.<br> Find the area shaded in green.
avanturin [10]

The area shaded in green is 864 cm²

<h3>Similar figures</h3>

Similar figures, corresponding angles are congruent and the sides are ratio of each other. Therefore,

AB / PQ = CD / RS

30 / 10 = 24 / RS

30RS = 240

RS = 240 / 30

RS = 8 cm

let find the height of trapezium PQRS.

AB / PQ  = 36 / h

30 / 10 = 36 / h

30h = 360

h = 360 / 30

h = 12 cm

Therefore,

area of the green portion = area of ABCD - area of PQRS

<h3>Area of a trapezium</h3>
  • area = 1 / 2(a + b)h

Therefore,

area of ABCD = 1 / 2(24 + 30)36 =  1 / 2 (54)36 = 1944 / 2 = 972 cm²

area of PQRS = 1 / 2(10 + 8)12 = 1 / 2(18)12 = 216 / 2 = 108 cm²

Area of the green portion = 972 - 108 = 864 cm²

learn more on trapezium here: brainly.com/question/11961445

6 0
2 years ago
Cobalt-56 has a decay constant of 8.77 × 10-3 (which is equivalent to a half life of 79 days). How many days will it take for a
sweet-ann [11.9K]
As I read this one, is just a decay exponential equation... so

A = P(1 + r)ᵗ   where "t" is days passed. . hmm in this case is decay, so negative rate  A = P(1 - r)ᵗ, and the decimal amount would be 0.00877 for the rate

62% of P, the original value, is just 0.62P, now... if we hmm take P as just 1, it could be any amount, but 62% of 1,000,000 is just 62% of 1 times 1,000,000

so, for the sake of comparing it with a percentage, 1 will do

\bf 0.62P=P(1-0.00877)^t\implies P=1\implies 0.62=(1-0.00877)^t&#10;\\\\\\&#10;ln(0.62)=ln[(1-0.00877)^t]\implies ln(0.62)=t\cdot  ln(1-0.00877)&#10;\\\\\\&#10;\cfrac{ln(0.62)}{ln(0.99123)}=t
7 0
3 years ago
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