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2 years ago

Which property is shown? If m∠ABC = m∠CBD, then m∠CBD = m∠ABC

Mathematics

2 answers:

goblinko [34]2 years ago

7 0

Answer: The answer is Symmetric Property

otez555 [7]2 years ago

6 0

ANSWER: symmetric property

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A teacher and a doctor each have 45 books. If 4/5 f the teacher's books and 2/3 of the doctor's books are novels, how many more

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Answer:

3/5

Step-by-step explanation:

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2 years ago

Write a number that has a hundreds digit that is greather than its tens digits

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Step-by-step explanation:

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3 years ago

Solve the equation.<br> 1/3(8x-9)=10x/3-(x+9)/6

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The answer to this question is x= -3

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3 years ago

Robert borrowed $750 from his parents. He promised to repay the loan by giving his parents at least $67 from his paycheck each w

Luda [366]

To answer this question we can propose the following equation:

$d=750 -x(67)$

Where:

x is the number of weeks elapsed.

d is the debt depending on the weeks.

Robert promised to pay at least $ 67 each week.

Therefore, after 5 weeks the debt of rober should be:

$d\leq750-5*67$

$d\leq415$

Then scenario A) is possible because $150$ .

After 6 weeks:

$d\leq750-6*67$

$d\leq348$

Then scenario B) is possible because $200$ .

After 7 weeks

$d\leq750-7*67$

$d\leq281$

Then scenario C) is possible because $250$ .

After 8 weeks

$d\leq750-8*67$

$d\leq214$

So scenario D) is NOT possible because $300>214$

Finally the correct answer is option D.

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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 $k\in\mathbb{R}$

(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)).

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3 years ago