Write an equation of the line that passes through the points (2,−1) and (0,1).

Blank divided by ( blank +blank ) =1

Allushta [10]

Answer:

3 divided by (1+2)=1

Step-by-step explanation:

bc u do parthensis first so 1+2 is 3 therfore if u divide 3 by 3 gives u one

4 0

3 years ago

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Which situation CANNOT be represented by this equation?

aniked [119]

Answer:

1

Step-by-step explanation:

8 0

4 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

What is the solution to the system of equations? (5, 0) (0, 5) (0, –5) (–5, 0)

adelina 88 [10]

The solution is the point where the lines intersect (cross)...and that would be :
(0,-5)

4 0

3 years ago

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The graph of a quadratic function is shown. Check all of the following statements that are true of the function.

Fed [463]

Answer:

a < 0

The vertex is (1,2)

Step-by-step explanation:

Looking at the graph

we have a vertical parabola open downward

The leading coefficient is negative

The vertex is a maximum

The vertex is the point (1,2)

The axis of symmetry is equal to the x-coordinate of the vertex, so the axis of symmetry is x=1

The y-intercept is the point (0,1)

The function has two real solutions (zeros of the function) one positive and one negative

The positive zero of the function is greater than 2

The negative zero of the function is greater than -1

Verify each statement

case 1) a < 0

The statement is true

Because the parabola open downward

case 2) The vertex is (1,2)

The statement is true

The vertex in this problem is the maximum of the function

case 3) The axis of symmetry is y =2

The statement is false

Because the axis of symmetry is x=1

case 4) x = 2 is a zero of the function

The statement is false

Because the positive zero of the function is greater than 2

case 5) (1,2) is a minimum

The statement is false

Because the vertex in this problem is a maximum

4 0

3 years ago