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1 year ago

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Determine whether a quadratic model exists for the set of values below. If so, write the model. f(0) = -4, f(3) = -19, f(-1) =

-11
Select the correct choice below:

A) A quadratic model does exist. It is f(x)= ?
B) A quadratic model does not exist.

1 answer:

Vikentia [17]1 year ago

7 0

Answer:

A) The model exists: f(x) = -3x^2 +4x -4

Step-by-step explanation:

A quadratic model will always exist for 3 given points, provided they are not on a line. In that case, a linear model is appropriate.

Here, the slope between -1 and 0 is positive, and the slope between 0 and 3 is negative. Thus, we know these points are not collinear, and a model must exist.

The model is most easily found using a quadratic regression tool. Such is shown in the attachment. It tells us that ...

f(x) = -3x^2 +4x -4

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I need a few answers<br> 1. P+15=-2 p=?<br> 2. 8y=-64 y=?<br> 3. -2 x 6 =?<br> 4. H-(-4) for H=8

Nikitich [7]

1: -17 + 15 = -2 P = -17
2: 8 x -8 = -64 y = -8
3: -2 x 6 = -12
4: 8 - (-4) = 12
Hope I Helped

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2 years ago

Find the population density of a city with an area of 598 square miles and a population of 21,500 people.

Pie

Answer: $35.12\text{ persons per square mile}$

Step-by-step explanation:

Given: The area of a city = 598 square miles

The population of city = 21,500 people.

The population density for a city is given by :-

$\text{Population density }=\frac{\text{Number of people }}{\text{Land area}}\\\\\Rightarrow\ \text{Population density }=\frac{21000}{598}\\\\\Rightarrow\ \text{Population density }=35.11705\approx35.12\text{ persons per square mile}$

Hence, the population density of a city = $35.12\text{ persons per square mile}$

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2 years ago

What is the value of x? 63, 42, 50

taurus [48]

Answer:

Step-by-step explanation:

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2 years ago

F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

$x^2-3x-4=0$

This implies that;

$x^2+x-4x-4=0$

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

8 0

2 years ago

Read 2 more answers

3n + 7 - 2<br> Combing like terms

hjlf

Answer:

Simplifying

3n + 7 + -2 = 0

Reorder the terms:

7 + -2 + 3n = 0

Combine like terms: 7 + -2 = 5

5 + 3n = 0

Solving

5 + 3n = 0

Solving for variable 'n'.

Move all terms containing n to the left, all other terms to the right.

Add '-5' to each side of the equation.

5 + -5 + 3n = 0 + -5

Combine like terms: 5 + -5 = 0

0 + 3n = 0 + -5

3n = 0 + -5

Combine like terms: 0 + -5 = -5

3n = -5

Divide each side by '3'.

n = -1.666666667

Simplifying

n = -1.666666667

Step-by-step explanation:

6 0

2 years ago