What are the like terms in the algebraic expression?

The graph of f(x) = x2 + x - 6 is shown below.

guapka [62]

Answer:

the solutions of a function are the points where for some value of x the function becomes zero

thus the solns for this graph would be

that's option 1.

3 0

2 years ago

7/2x+1/2=10 1/2+9/2x what is the value of x

gogolik [260]

Answer:

21

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Bert has three times as much money as Ernie. If Ernie spends $6 on a new rubber ducky, then Bert will have five times as much mo

fomenos

Answer:

Bert has $45.

Step-by-step explanation:

I have no real strategy, except for that I used guess and check. How did I do this? Well, since I am in 5th Grade, I don't know algebra very well, so I made an organized chart and checked all the numbers that had to be a multiple and could be divided equally by 5, 3 and when a third of that number was subtracted by 6, it was a fifth of the orginal number. Thats how I got 45.

Checking this answer:

It is always important to check your answer after finishing the problem, so this is how I checked my answer:

1. 45 divided by 3 = 15

2. 15 - 6 = 9, and 9 is 1/5 of 45

Bert has $45.

5 0

3 years ago

3×52 A. 5/6 B. 15/2 C.8/2 D.15/6

ohaa [14]

D.) 15/6

- 3 times 52 = 156, and since you can't exactly convert it to a fraction then i'm assuming that it is 'D'.

- Ouma :>

6 0

3 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