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2 years ago

Witch expression is equivalent to 1/5(150x - 80y + 50 - 50x- 25y + 20)?

Mathematics

1 answer:

Advocard [28]2 years ago

7 0

Answer:

20x - 21y + 14

Step-by-step explanation:

1/5(150x - 80y + 50 - 50x- 25y + 20)

Step 1 - Distribute

1/5 • 150x = 30x

1/5 • (-80y) = -16y

1/5 • 50 = 10

1/5 • (-50x) = -10x

1/5 • (-25y) = -5y

1/5 • 20 = 4

Step 2 - Combine Variables

30x - 10x = 20x

-16y - 5y = -21y

10 + 4 = 14

Step 3 - Arrange the Set of Numbers

20x - 21y + 14

This is the simplified expression, I know it is correct, I doubled checked on my expression calculator.

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If a car travels at an average speed of x miles per hour, how far would the car travel in 90 minutes ?

antiseptic1488 [7]

Answer:

(1.5)x

Step-by-step explanation:

90 minutes = 1.5 hours therefore

if car travels x miles in 1 hour

then the care has to travel (1.5)x miles in 90 minutes

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3 years ago

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Identify the binomial that is a factor of the polynomial P(x) = 2x^3 + 12x^2 − 26x − 84.

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2 years ago

Answer please. Simplify the expression 8y - 11y + 6

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8y - 11y + 6

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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)).

4 0

3 years ago

What is the slope Of the line that passes through the points in the picture?

GuDViN [60]

Answer:

5/3

Step-by-step explanation:

To find the slope we pick one set of points

2/3,-1/2

Set your equation up like this

y2-y1/x2-x1

Plug in and subtract(notice the negative)

And we get 5/3

Hope this helps

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2 years ago