What are the factors of x2-x-6? Check all that apply X-6 X+6 X-3 Ox-2 Ox+2

Mathematics

1 answer:

snow_lady [41]3 years ago

7 0

Answer:

f(x) = x^2 - x - 6 = x^2 + 2x - 3x - 6 = x(x + 2) - 3(x + 2) = (x - 3)(x + 2)

=> x - 3 and x + 2 are factors of f(x).

Hope this helps!

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76 cm a 52 cm Find the area of the regular pentagon (the answer has one digit after the decimal point)

alekssr [168]

Answer:

98.8 cm²

Step-by-step explanation:

Hi there!

$A=\frac{1}{2} Pa$ where P is the perimeter of the pentagon and a is the apothem (the line going from the center of the pentagon one edge)

1) Find the perimeter

$P=5l$ where l is the side length

Plug in the side length 7.6 cm

$P=5(7.6)\\P=38$

Therefore, the perimeter of the pentagon is 38 cm.

2) Find the area

$A=\frac{1}{2} Pa$

Plug in the perimeter (38) and the apothem (5.2)

$A=\frac{1}{2} (38)(5.2)\\A=\frac{1}{2} (38)(5.2)\\A=\frac{1}{2} (197.6)\\A=98.8$

Therefore, the area of the pentagon is 98.8 cm².

I hope this helps!

5 0

3 years ago

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$