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lukranit [14]
3 years ago
12

Carl is deriving the quadratic equation by completing the square. His work at one of the steps is shown. x + b/na = + or - b^2-4

ac. Carl is not sure what the value of n is. What value should he use to continue deriving the quadratic equation correctly? A. 1 B. 2 C. 3 D. 4
Mathematics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

the correct option is B. 2

Step-by-step explanation:

As given,

x + \frac{b}{na} = ± √b²-4ac

⇒x = -\frac{b}{na} ± √b²-4ac

As e know that the quadratic formula is

x = \frac{-b + or - \sqrt{b^{2} - 4ac }  }{2a}

So , by comparing with the formula

we get

n = 2

So, the correct option is B. 2

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Which method of solving quadratic equations would be the easiest to use to find the solution for x2 – 13x + 12 = 0?
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A council of three people is to be chosen from a group of 4 lawyers, one artist, and 3 teachers
padilas [110]

Answer: 0.2857

<u>Step-by-step explanation:</u>

At least one teacher means exactly 1 or exactly 2 or exactly 3 teachers. Find the probability of each and add them up.

\text{One teacher}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{5\ others}{7\ people}\times \dfrac{4\ others}{6\ people}=\dfrac{5}{28}=0.1786\\\\\\\text{Two teachers}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{2\ teachers}{7\ people}\times\dfrac{5\ others}{6\ people}=\dfrac{5}{56}=0.0893\\\\\\\text{Three teachers}:\dfrac{3\ teachers}{8\ total\ people}\times\dfrac{2\ teachers}{7\ people}\times\dfrac{1\ teacher}{6\ people}=\dfrac{1}{56}=0.0179

One teacher + Two teachers + Three teachers = 0.2857

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3 years ago
Drag each percent to match the situation. Each percent may be used only once.
Svetllana [295]

Answer:

1:80%

2:95%

3:15%

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7 0
3 years ago
A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
VMariaS [17]

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

                   -F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

                    F_n,b - m*g = 0

                    F_n,b = m*g

- The frictional force F_f is given by:

                   F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

                  - m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

                  4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

                        6.6*cos(Q) = 4*u_s*sin(Q)

                               tan(Q) = 6.6 / 4*u_s

- Plug in the values:

                               tan(Q) = 6.6 / 4*0.23

                                    Q = arctan(7.1739) = 82.06

                   

                     

4 0
3 years ago
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