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3 years ago

Help me please thanks

Mathematics

2 answers:

Nadya [2.5K]3 years ago

8 0

Answer: (D) all real numbers

Step-by-step explanation:

4(a + 3) = 12 + 4a

4a + 12 = 12 + 4a distributed 4 into a + 3

12 = 12 subtracted 4a from both sides

TRUE

A true statement means they are the same equation so any value you plug in for "a" will make a true statement. There are an infinite number of solutions.

Examples:

Let a = 0, then 4(0 + 3) = 12 + 4(0)

⇒ 4(3) = 12 + 0

⇒ 12 = 12

Let a = 10, then 4(10 + 3) = 12 + 4(10)

⇒ 4(13) = 12 + 40

⇒ 52 = 52

Let a = 100, then 4(100 + 3) = 12 + 4(100)

⇒ 4(103) = 12 + 400

⇒ 412 = 412

Any number you choose, will make a true statement so the solution is "all real numbers".

Natalka [10]3 years ago

6 0

Answer: All real numbers

Step-by-step explanation: If you plug in any number, it will work. 4(a+3) = 12+4a. a=-3 0=0

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Step-by-step explanation:

The center of measure calculated was the average or mean. To get the mean of a set of numbers, we have to add all the numbers given and then divide them by the numbers.

In this case, the mean of 3,7,11,11,16 will be:

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3 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

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Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is a separable differential equation. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

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