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2 years ago

Which shows the factored form of x2 – 12x – 45?

2 answers:

6 0

Hi!

(x+a)·(x+b) = x²+xa+xb+ab = x²-12x-45

x² = x²
xa+xb = (a+b)x = -12x ⇒ (a+b) = -12
ab = -45

When (a+b) = -12 and ab = -45?

+3-15 = -12 and (+3)(-15) = -45
-3-15 = -18 and (-3)(-15) = +45
+3+15 = +18 and (+3)(+15) = +45
-3+15 = +12 and (-3+15) = -45

Answer:

(x+3)(x-15)

elixir [45]2 years ago

4 0

Answer:

Howdy ya'll!!!

Step-by-step explanation:

I know that the answer would be (x+3) and (x-15). Good luck on whatever you are doing!!

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On a coordinate plane, triangle A B C is shown. Point A is at (3, 4), point B is at (negative 5, negative 2), and point C is at

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Answer:

The perimeter of Δ ABC is 20 + 2 $\sqrt{10}$ units ⇒ Last answer

Step-by-step explanation:

The perimeter of any triangle is the sum of the lengths of its three sides

The formula of distance between two points is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

In Δ ABC

∵ A = (3 , 4) , B = (-5 , -2) , C = (5 , -2)

∵ AB = 10 units

∵ AC = 2 $\sqrt{10}$

- To find its perimeter find the length of BC

∵ $x_{1}$ = -5 and $y_{1}$ = -2

∵ $x_{2}$ = 5 and $y_{2}$ = -2

- By using the formula above

∴ $BC=\sqrt{(5--5)^{2}+(-2--2)^{2}}=\sqrt{(5+5)^{2}+(-2+2)^{2}}$

∴ $BC=\sqrt{(10)^{2}+(0)^{2}}=\sqrt{100}$

∴ BC = 10 units

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∵ P = AB + BC + AC

∴ P = 10 + 10 + 2 $\sqrt{10}$

- Add like terms

∴ P = 20 + 2 $\sqrt{10}$

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A fitness club holds an open house for new members. The club expects 6,500 people to sign up for gym membership. If there is a p

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F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

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Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

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$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

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2 years ago

Which transformations have been applied to the graph of f(x) = x2 to produce the graph of g(x) = –5x2 + 100x – 450? Select three

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Answer:

The graph of f(x) is shifted up 50 units

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Step-by-step explanation:

we have

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This is a vertical parabola open upward

The vertex is a minimum

The vertex is the origin (0,0)

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This is a vertical parabola open downward

The vertex is a maximum

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Find the vertex of g(x)

Convert to vertex form

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Complete the square

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$g(x)=-5(x^{2}-20x+100)+50$

$g(x)=-5(x-10)^{2}+50$

The vertex is the point (10,50)

To translate the vertex of (0,0) to (10,50)

The rule of the translation is

(x,y) ------> (x+10,y+50)

That means ----> The translation is 10 units at right and 50 units up

therefore

The transformations are

The graph of f(x) is shifted up 50 units

The graph of f(x) is shifted right 10 units

The graph of f(x) is reflected over the x-axis

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