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JulsSmile [24]
2 years ago
13

An x-method chart shows the product a c at the top of x and b at the bottom of x. Above the chart is the expression a x squared

b x c. Which statement is true of x2 8x – 6? (x – 2) is a factor of the polynomial. (x 2) is a factor of the polynomial. (x – 4) is a factor of the polynomial. The polynomial is prime.
Mathematics
1 answer:
bearhunter [10]2 years ago
5 0

The polynomial is prime because the provided polynomial can not be factored shown in the X-method chart.

<h3>What is X method?</h3>

X-method is the method to find the factors of the polynomial equation.

An x-method chart shows the product a c at the top of x and b at the bottom of x.

Above the chart is the expression,

ax^2+bx+c

The given polynomial in the problem is,

x^2 +8x - 6

Let's check if the factors of polynomial. To find the factor, we need, to find the two number whose sum or different is 8 and the product is 6 (6 × 1).

As there is no number which gives has the sum or different is 8 and the product is 6 (6 × 1). Thus, the above polynomial can not be factored. Thus this polynomial is prime.

Thus, the polynomial is prime because the provided polynomial can not be factored, shown in the X-method chart.

Learn more about the X method here;

brainly.com/question/24379507

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Identify the x-intercept and y-intercept of the line 4x-2y=-12
Alex73 [517]
To find x-intercept, put the Y value = 0;

4x - 2y = -12

-> 4x -2.0 = -12
-> 4x = -12
-> x = -12/4
-> x = -3

X-intercept: (-3,0)

Now do the reverse to find the y-intercept, X = 0;

4x - 2y = -12

-> 4.0 - 2y = -12
-> -2y = -12 x(-1)
-> 2y = 12
-> y = 12/2
-> y = 6

Y-intercept: (0, 6)
7 0
3 years ago
How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

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Sara weight 25 pound-more than amber if together they weight 205 what is the weight of sara ??
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20 minus the product of 3 and 5.
olganol [36]

Answer:

5

Step-by-step explanation:

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