2<br> Which of the following is equivalent to the expression<br> : 4?<br> 5

Alan has forgotten his 4-digit PIN code.

SVEN [57.7K]

Answer:

a lot of different answers

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The graph of y = f(x) has a maximum point at (-2, 4).

solmaris [256]

Answer:

(2,4).

Step-by-step explanation:

Graph of y = f(-x).

The values of x in f(-x) assume the values of -x in f(x).

For example, f(-2) in f(x) is f(2) in f(-x), and f(2) in f(x) is f(-2) in f(-x).

b) Write down the coordinates of the maximum point of the graph of y = f(-x).

In f(x), we have that x = -2. So in f(-x), it will be x = 2. y remains unchanged. So

(2,4).

5 0

3 years ago

I need help solving -23<1-4x-8

Elan Coil [88]

Answer is x is less than 4

7 0

3 years ago

Where the perimeter of the base is 7.1 cm and the height of the pyramid is 10.8 cm

xz_007 [3.2K]

Answer:

17.9cm

Step-by-step explanation:

Its is very easy you just add 7.1 to 10.8.

Perimeter you add

Area you multiply

4 0

2 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

2 Which of the following is equivalent to the expression : 4? 5