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Scrat [10]
3 years ago
13

George is helping the manager of the local produce market expand her business by distributing flyers

Mathematics
1 answer:
spin [16.1K]3 years ago
4 0

Answer:

20 + 0.05x ≥ 65

Step-by-step explanation:

20 is by itself since thats what he earns extra per day.

Since he earns 0.05 per flyer, that means we are going to need a variable since we don't know how many flyers he passes. Let x be equal to that.

Since he wants to make at LEAST 65 we are going to use the greater than or equal to symbol since he doesn't want to make less than that.

20 + 0.05x ≥ 65

Best of Luck!

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Plz help me. But like for real plz
scoundrel [369]

Answer:

They will pay the same amount!

3 0
2 years ago
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Solve for x: 5x + 15 = 75
yuradex [85]

Answer:

x = 12

Step-by-step explanation:

5x + 15 = 75

      -15   -15

5x = 60

x = 12

3 0
2 years ago
How to solve x=6y-11 and 3x-2y=-1 by substitution
vagabundo [1.1K]
Substitute x into the second equation:
3(6y-11)-2y=-1
18y-33-2y=-1
16y=32
y=2
Then put the y value into the first equation to get x:
x=6(2)-11
x=1
There you go!

Please mark brainliest
7 0
2 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
Andrews [41]

Answer:

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

Step-by-step explanation:

This is <em>a separable differential equation</em>. 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)

4 0
2 years ago
HELP!! ASAP!!!!!!!!!!!!!!!!!!!1
Artemon [7]

Answer:

No, No, Yes, Yes

Step-by-step explanation:

6 0
3 years ago
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