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Allushta [10]
3 years ago
11

Does(3, 6)make the inequality x + y ≤ 9 true?

Mathematics
2 answers:
mario62 [17]3 years ago
7 0
Hi


(x,y) = (3,6) ⇒ x=3 and y=6

x+y ≤ 9
3+6 ≤ 9
9 ≤ 9 ⇒ 9 is lower than or equal than 9? Yes, true, 9 = 9

Answer: (3,6) is a solution
maw [93]3 years ago
6 0
If you would like to know if (3, 6) is a solution to the inequality x + y <= 9, you can calculate this using the following steps:

(3, 6) ... x = 3, y = 6
x + y <= 9
3 + 6 <= 9
9 <= 9

Result: (3, 6) is a solution to the inequality x + y <= 9.
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This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

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                               \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

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

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      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.

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Hence,

h(x) =8x^3-19x^2+14x-3

Keywords: Functions, product

Learn more about functions at:

  • brainly.com/question/8054589
  • brainly.com/question/7932185

#LearnwithBrainly

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