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pychu [463]
3 years ago
15

Complete the solution of the equation. Find the value of y when x equals -18. -3x - y = 56

Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer:

y = -2

Step-by-step explanation:

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The change in water vapor in a cloud is modeled by a polynomial function, C(x). C(x) = x^3+ 5x^2+3x+30 Describe how to find the
kari74 [83]

Answer:

Suppose the equation that describes the change in water vapor in a cloud (y) with respect to temperature (x): C(x) = x³ - 4x² - 7x + 10

You find the x-intercepts of the cubic equation by finding the roots. Let y be zero:

x³ - 4x² - 7x + 10 = 0

Now, find possible factors of the constant term in the equation: 10. These could be 1, 2, 5, and 10. Suppose we choose 1. We substitute x=1. If the answer is zero, then x=1 is one of the roots.

(1)³ - 4(1)² - 7(1) + 10 = 0

Hence, x=1 or x-1 = 0 is one of the roots. We use this to manipulate the rest of the terms in the original equation, such that you can factor out (x-1). Substitute -x² - 3x² to -4x³ because they are just equal. Same is true for +3x - 10x for -7x:

x³ -x² - 3x²+3x - 10x + 10 = 0

Factor out like terms such that they have a common factor of (x-1) as much as possible.

x³ -x² = x²(x-1)

- 3x²+3x = -3x(x-1)

- 10x + 10 = -10(x-1)

The factored equation is:

x²(x-1)-3x(x-1)-10(x-1) = 0

Factor out (x-1):

(x-1)(x²-3x-10) = 0

Now, we have the quadratic equation left to be solved:x²-3x-10

Use the quadratic formula to find the roots:

x = [-b +/- √(b²-4ac)]/2a,

The a, b and c coefficients are based on the general form of the quadratic equation: ax² + bx + c. Hence, a=1, b=-3 and c=-10. Substituting the values:

x = [-(-3) +/- √((-3)²-4(1)(-10))]/2(1)

x = 5, -2

Therefore, the roots of the cubic equations are x = 1, x = 5 and x =- 2. These are the x-intercepts of the equation. At temperatures °C, 5°C and -2°C, there is no change in water vapor in a cloud. The graph is shown in the attached picture

Step-by-step explanation:

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3 years ago
9×(-1/3)×(-3)×(-1/9)=?​
bulgar [2K]

Answer:

-1

Step-by-step explanation:

7 0
3 years ago
Triangle PQR has coordinates P(–8, 3), Q(–8, 6), and R(–3, 6). If the triangle is translated by using the rule (x, y) right-arro
amid [387]

Answer:

Step-by-step explanation: in the x's place add 4 to that number and in the y's place subtract 6 from that number:

P= -4,-3

Q= -4,0

R= 1, 0

3 0
3 years ago
Read 2 more answers
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
3 years ago
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