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Helga [31]
3 years ago
5

Find all the real zeros of f(x)=3x^3-9x^2+3x-9 Can someone help me please

Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

x=0,3

Step-by-step explanation:

start by dividing every term by 3

x^3-3x^2+x-3

group into 2 terms

(x^3-3x^2)+(x-3)

simplify as much as you can

x^2(x-3)+(x-3)

combine terms

(x^2)(x-3)

x=3, 0

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Gabe rolled 14 strikes out of 70 attempts. What percent of Gabe's attempts were strikes?
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Answer:

20%

Step-by-step explanation:

14/70x100=20%

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3 years ago
PLEASE HELP ILL GIVE BRAINLIEST
dusya [7]

1)

(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}

3)

let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}

7 0
3 years ago
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When the angle of elevation to the sun is 52", a tree casts a shadow that is 13 feet long. What is
shepuryov [24]

Answer:

72

Step-by-step explanation:

6 0
3 years ago
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What is the equation of the quadratic graph with a focus of (4, −3) and a directrix of y = −6?
Artyom0805 [142]

Answer:

y=\frac{1}{6} (x-4)-\frac{9}{2}

Step-by-step explanation:

∵ The quadratic equation form is :

y = [1/2(b - k)] (x - a)² + (b + k)/2

Where (a , b) is the focus and directrix y = k

∵ The focus is (4 , -3) and directix is y = -6

∵ y=\frac{1}{2(-3-(-6))} (x-4)^{2}+\frac{(-3)+(-6)}{2}

∴ y=\frac{1}{6} (x - 4)^{2}+\frac{-9}{2}

∴ y=\frac{1}{6} (x-4)-\frac{9}{2}

another way:

Assume that (x , y) is the general point on the parabola

∵ The distance between the directrix and (x , y) = the distance between the focus and (x , y)

By using the distance rule:

∵ (y - -6)² = (x - 4)² + (y - -3)² ⇒ (y + 6)² = (x - 4)² + (y + 3)

∴ y² + 12y + 36 = (x - 4)² + y² + 6y + 9

∴ 12y - 6y = (x - 4)² + 9 - 36

∴ 6y = (x - 4)² - 27 ⇒ ÷ 6

∴ y = 1/6 (x - 4)² - 9/2

4 0
3 years ago
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Only the upper-left statement is NOT true.
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