What are the zeros of the polynomial equation f(x)=x^2-12x+20?

1 answer:

8 0

$f(x)=x^2-12x+20\\\\\text{The zeros:}\\\\f(x)=0\to x^2-12x+20=0\\\\x^2-10x-2x+20=0\\\\x(x-10)-2(x-10)=0\\\\(x-10)(x-2)=0\iff x-10=0\ \vee\ x-2=0\\\\\boxed{x=10\ \vee\ x=2}$

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Pls help asap whats the local min value of the function below?

shutvik [7]

Answer:

given function has 2 minimums - $\frac{9}{4}$ and $\frac{9}{4}$

Step-by-step explanation:

Step 1. g'(x) = 4x³ - 10x

Step 2. Find find the critical points:

4x³ - 10x = 2x(2x² - 5) = 0

$x_{1}$ = - $\sqrt{\frac{5}{2} }$ , $x_{2}$ = 0 , $x_{3}$ = $\sqrt{\frac{5}{2} }$

Step 3. g'(x) > 0 : - $\sqrt{\frac{5}{2} }$ < x < 0 or x > $\sqrt{\frac{5}{2} }$

g'(x) < 0 : x < - $\sqrt{\frac{5}{2} }$ or 0 < x < $\sqrt{\frac{5}{2} }$

Step 4.

If x ∈ ( - ∞ , - $\sqrt{\frac{5}{2} }$ ) , g(x) is decreasing ;

If x = - $\sqrt{\frac{5}{2} }$ , g(x) has minimum value ;

If x ∈ ( - $\sqrt{\frac{5}{2} }$ , 0 ) , g(x) is increasing ;

If x = 0 , g(x) has maximum value ;

If x ∈ ( 0 , $\sqrt{\frac{5}{2} }$ ) , g(x) is decreasing ;

If x = $\sqrt{\frac{5}{2} }$ , g(x) has minimum value ;

If x ∈ ( $\sqrt{\frac{5}{2} }$ , ∞ ) , g(x) is increasing .

⇒ at ( - $\sqrt{\frac{5}{2} }$ , - $\frac{9}{4}$ ) and at ( $\sqrt{\frac{5}{2} }$ , $\frac{9}{4}$ ) , g(x) reaches its minimum

8 0

3 years ago

Carla's plant was 10 2/3 inches tall. Yolanda's was 8 3/4 inches tall.

marin [14]

10 2/3 inches=32/3 inches
8 3/4 inches= 35/4 inches

Find the common denominator, in this case, 12.

128/12 inches is Carla's plant

105/12 inches is Yolanda's plant.

128/12 > 105/12. Therefore, Carla's plant is taller than Yonlanda's

3 0

3 years ago

Events A and B are such that P(A) = 3 8 , P(B) = 2 7 and P(A ∩ B) = 1 7 Work out P(A|B)

tigry1 [53]

Answer:

$\frac{17}{27}$