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

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:

7 0

3 years ago

HELP PLEASE, Part a of the question was posted earlier on.

Elanso [62]

$\bf \lim\limits_{x\to 6}~\cfrac{x^2+x+42}{x-6}\implies \lim\limits_{x\to 6}~\cfrac{(x+7)(x-6)}{(x-6)}\implies \lim\limits_{x\to 6}~x+7
\\\\\\
\lim\limits_{x\to 6}~(6)+7=13
\\\\\\
\textit{since the original equation holds for all, }x\ne 6,\textit{ then it approaches}\\
\textit{the same limit as }x+7$

3 0

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 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.

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