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blondinia [14]
3 years ago
13

Between which two ordered pairs does the graph of f(x) = one-halfx2 + x – 9 cross the negative x-axis?

Mathematics
2 answers:
Anit [1.1K]3 years ago
6 0

Answer:

Step-by-step explanation:

we have that

f(x) = x^2 + x - 9

Using a graph tool

see the attached figure

The function represent a vertical parabola that open up

the vertex is a minimum-------> is the point (-0.5, -9.3)

The x-intercepts are the points when the y coordinate is equal to zero

The x-intercepts are the points (-3.5, 0)  and (2.5, 0)

so

the function cross the negative x-axis at point

therefore, the answer is

(-4, 0) and (-3, 0)

k0ka [10]3 years ago
3 0

Answer:

(-6,0)(-5,0)

Step-by-step explanation:

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Answer:

\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}

Step-by-step explanation:

The <u>width</u> of a square is its <u>side length</u>.

The <u>width</u> of a circle is its <u>diameter</u>.

Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.

<u>Formulas</u>

\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}

\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}

\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}

If the diameter is equal to the side length of the square, then:
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Therefore:

\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}

So the ratio of the area of the circle to the original square is:

\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}

Given:

  • side length (s) = 6 in
  • radius (r) = 6 ÷ 2 = 3 in

\implies \sf \textsf{Area of square}=6^2=36\:in^2

\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)

Ratio of circle to square:

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