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worty [1.4K]
3 years ago
11

2. What are the zeroes of P(x) = x3 – 6x2 – x +62

Mathematics
1 answer:
Inga [223]3 years ago
3 0

Step-by-step explanation:

if u mean (x^3 -6x^2 -x- 6) then the answer is C

(6,-1,1)

..................,..........

but if u mean P(x) = x3 – 6x2 – x +62

then the answer is None of the above

-2.724

4.361+1.9328i

4.361+1.9328i

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Khan academy question
andriy [413]

Answer:

(2,5)

Step-by-step explanation:

There you go.........

4 0
3 years ago
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Please help on this math problem!!!! will mark brainliest!!!!
worty [1.4K]

Answer:

\frac{14}{33}

Step-by-step explanation:

1/6 + 1/6 + 1/3 + 1/3 + 1/3 + 1/2 + 2/3 + 2/3 + 2/3 + 5/6 = 14/3

(14/3)/11 = 14/33

Hope this helps!

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4 0
3 years ago
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Encuentra la ecuacion de la parabola con vertice en el origen y foco en las cordenadas 0,1en su forma ordinaria y general
marusya05 [52]

Answer:

x^{2}=4y

y=\frac{1}{4}x^{2}

Step-by-step explanation:

Forma ordinaria

La ecuación de la parabola de manera ordinaria está dada por:

(x-h)^{2}=4p(y-k) (1)  

Donde:

  • (h,k) es la coordenda del vértice, en nuestro caso (0,0) ya que está en el origen.
  • (h,k+p) es la coordenda del foco, en nuestro caso (0,1).

Por lo tanto h = 0, k = 0 y p = 1.

Remplazando estos valores en la ecuación de la parábola, tenemos:

(x-0)^{2}=4*1(y-0)

x^{2}=4y (2)

 

Forma general  

La forma general de una parábola esta dada por la siquiente ecuación

y=ax^{2}+bx+c

Reordenando la ecuación (2) tenemos:

y=\frac{1}{4}x^{2}

Espero esto te haya ayudado!

5 0
3 years ago
What is the rule? 100 cm 1 m, 250 cm ? M, 1500 cm ? M
kirza4 [7]
There are 100 cm(centimeters) in a m(meter).
So, to solve this, we divide 250 by 100. this gives us 2.5meters.
1500÷100=15meters
The rule is m=cm÷100.

4 0
3 years ago
The rectangular bar is connected to the support bracket with a 10-mm-diameter pin. The bar width is w = 70 mm and the bar thickn
german

Answer:

P_max = 9.032 KN

Step-by-step explanation:

Given:

- Bar width and each side of bracket w = 70 mm

- Bar thickness and each side of bracket t = 20 mm

- Pin diameter d = 10 mm

- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa

- Average allowable shear stress of pin S = 115 MPa

Find:

The maximum force P that the structure can support.

Solution:

- Bearing Stress in bar:

                                       T = P / A

                                       P = T*A

                                       P = (120) * (0.07*0.02)

                                       P = 168 KN

- Shear stress in pin:

                                        S = P / A

                                        P = S*A

                                        P = (115)*pi*(0.01)^2 / 4

                                        P = 9.032 KN

- Bearing Stress in each bracket:

                                       T = P / 2*A

                                       P = T*A*2

                                       P = 2*(120) * (0.07*0.02)

                                       P = 336 KN

- The maximum force P that this structure can support:

                                      P_max = min (168 , 9.032 , 336)

                                      P_max = 9.032 KN

3 0
3 years ago
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