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3 years ago

How do i solve 23.6-2x3/2

Mathematics

2 answers:

Vlada [557]3 years ago

4 0

Answer:

103/5 in fraction form or 20.6 in decimal form

Step-by-step explanation:

First, you follow pemdas. Multiplication and division first. 2 times 3 is 6. 6 divided by 3 is 3. 23.6 subtracted by 3 would equal 20.6.

White raven [17]3 years ago

3 0

Answer:

20.6

Step-by-step explanation:

23.6-2x3/2

Solve:

23.6-2x3/2

=23.6 - 6/2

= (47.2-6)/2

=20.6

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Answer:

12,8

Step-by-step explanation:

Just line them up and you will arrive at your answer:

5,8

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3 years ago

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AnnyKZ [126]

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3 years ago

Find the x-intercept of the parabola of with vertex (1,20) and the y-intercept (0,16). write your answer in this form: (x1,y1),(

svetoff [14.1K]

I assume that the parabola in this particular problem is one whose axis of symmetry is parallel to the y axis. The formula we're going to use in this case is (x-h)2=4p(y-k). We know variables h and k from the vertex (1,20) but p is not given. However, we can solve for p by substituting values x and y in the formula with the y-intercept:

(0-1)^2=4p(16-20)

Solving for p, p=-1/16.

Going back to the formula, we can finally solve for the x-intercepts. Simply fill in variables p, h and k then set y to zero:

(x-1)^2=4(-1/16)(0-20)
(x-1)^2=5
x-1=(+-)sqrt(5)
x=(+-)sqrt(5)+1

Here, we have two values of x

x=sqrt(5)+1 and
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thus, the answers are: (sqrt(5)+1,0) and (-sqrt(5)+1,0).

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3 years ago

5x-5=3x +17<br> what is the answer?

Ratling [72]

Answer:

5x-5=3x +17 So were solving for x The answer would be =11

Step-by-step explanation:

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3 years ago

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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