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2 years ago

Help me somebody pleaseee

Mathematics

1 answer:

alina1380 [7]2 years ago

5 0

Answer:

Hope i helped but i am not sure chief if that is what you needed

Step-by-step explanation:

you said in the comment we need to find the gallons so if we have the cost 10$ per 2 gallons then 20$ per 4 gallons. Then technicly 1 gallon will be 0 $

But if we need to find the value of the red dots it's starting from the top

40$ for the value of 8

30$ for the value for 6

20 $ for the value of 4

10$ for the value of 2

You can conclude that the cost bar increases by 2 each time

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The Sunny Days Apartment building has a cylindrical, above ground pool in the yard. The radius of the pool is 2 times its height

garik1379 [7]

9514 1404 393

Answer:

6.2 ft

Step-by-step explanation:

The volume of a cylinder is given by ...

V = πr²h

Here, we have r=2h and V=3000 ft³. Putting these numbers into the formula and solving for h, we get ...

3000 ft³ = π(2h)²h

(3000 ft³)/(4π) = h³ . . . . . divide by the coefficient of h³

h = ∛(750/π) ft ≈ 6.2 ft

The pool is about 6.2 feet deep.

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2 years ago

The function f(x)=30+0.25x/x yields the average cost in dollars per cup of lemonade made by Lincoln at his lemonade stand when h

trapecia [35]

The answer is C: Lincoln uses 25 cents worth of supplies per cup.

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3 years ago

Read 2 more answers

Is the phrase ( Moving back 5 spaces on a game board) negative or positive

Mkey [24]

Answer:

The phrase is negative because you're moving backwards

8 0

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

3 5 8m c1m 0 8 1m

miv72 [106K]

Answer:

:00000000000 DANGGGGGGGGGGGGGGGGGGGGGGG

Step-by-step explanation:

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3 years ago