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2 years ago

What is 3/x+6=x/9 ? add

Mathematics

1 answer:

Mkey [24]2 years ago

4 0

Let's solve your equation step-by-step.

3x+6=x9

Step 1: Cross-multiply.

3x+6=x9

(3)*(9)=x*(x+6)

27=x2+6x

Step 2: Subtract x^2+6x from both sides.

27−(x2+6x)=x2+6x−(x2+6x)

−x2−6x+27=0

Step 3: Factor left side of equation.

(−x+3)(x+9)=0

Step 4: Set factors equal to 0.

−x+3=0 or x+9=0

x=3 or x=−9

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is it possible for a number to be rational number that is not an integer is not an integer but is a whole number?explain.

Lostsunrise [7]

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3 years ago

Determine the # of X intercepts from this equation F(x)=x^2+5x+4 ?

k0ka [10]

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3 years ago

The height 'h' (in feet) of a ball in a baseball game can be modeled by h = -16t² + 28t + 8 , where 't' is the time (in seconds)

lidiya [134]

Answer:

a) No. t < 0 is not part of the useful domain of the function

b) 2.0 seconds

Step-by-step explanation:

a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.

b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.

h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0

0 = -4(4t² -7t -2) = -4(4t +1)(t -2)

This has t-intercepts where the factors are zero, at t=-1/4 and t=2.

The ball will hit the ground after 2 seconds.

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3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by