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2 years ago

The main goal of a personal budget is to

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:

have income be greater than expenses

Step-by-step explanation:

Evgesh-ka [11]2 years ago

3 0

Make sure you don't spend to much and have no more money left

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Given the following functions f(x) and g(x), solve (f/g)(-4) and select the correct answer below. F(x)=4x-4 G(x)=x-1

yan [13]

To start set up a fraction with the f(x) on top and g(x) on bottom
(f/g)(x) = (4x - 4)/(x - 1) - This is the function that we are going to use

Plug in -4 for x
(f/g)(-4) = (4(-4) - 4)/(-4 - 1) = (-16 - 4)/(-5) = (-20)/(-5) = 4

So...
(f/g)(-4) = 4

4 0

2 years ago

Read 2 more answers

What is the equation for X+3y=16;(-3,-4)

Galina-37 [17]

I hope this helps you

-3+3. (-4)=16

-3-12=16

-15=16 false

7 0

3 years ago

Ik it’s 7 in the morning but can somebody help a gal out

faust18 [17]

Answer:

Step-by-step explanation:

The vertical scale gives us the frequency or the number of pets under each category . From the diagram;

The number of Dogs is 11 while there is only 1 horse

The difference between this numbers will be the solution to the question posed;

11 - 1 = 10

Therefore, 10 more dogs are pets than horses

4 0

3 years ago

Read 2 more answers

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

Find the missing factor B that makes the equality true.-35^6= (-5x^2)(b)

Ber [7]

This is the answer!!!!

7 0

2 years ago