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Vikentia [17]
2 years ago
10

Combine like terms to simplify 4y+ 6x-2y

Mathematics
1 answer:
ladessa [460]2 years ago
3 0

Answer:

2y + 6x

Step-by-step explanation:

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Can anyone help me, i will give brainiest.
laila [671]

Answer: ∠DOB: 48°

Step-by-step explanation:

1. we need an equation first. the sum of all angles (108°, n°, 2n°) is equal to 180°. we can depict this with the equation: 108°+2n°+n°=180°

2. now we can solve for the missing variable, n.

108°+3n°=180° → subtract both sides by → 3n°=72° → divide both sides by 3 → n=24°

3. now that we know that n=24°, we can solve the value of ∠DOB. we can see that ∠DOB is 2n° which we just plug the number we got for n into the equation. 2*24=48° meaning ∠DOB is 48°

hope this heped! ♡

7 0
2 years ago
Read 2 more answers
What is the domain and range for this graph?
Zinaida [17]

<u>Domain</u> is the <u>x value</u> (here,y) while the <u>range</u> <u>value</u> is <u>y</u> (x,here).

The image is blurry for me but I have explained how to complete this problem, so it should be easier for you.  

3 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
Can someone help me please?
blagie [28]

the abswer of this question is c

6 0
3 years ago
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Two triangles are said to be congruent, if three sides of one triangle are equal to the three corresponding sides of another tri
Natasha2012 [34]

Answer:

Side-Side-Side (SSS) Rule

8 0
3 years ago
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