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zaharov [31]
3 years ago
11

The table below represents a linear function f(x) and the equation represents a function g(x):

Mathematics
1 answer:
Masja [62]3 years ago
8 0

Answer:

what is linear

Step-by-step explanation:

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What is the solution of the equation (x – 5)^2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.
xz_007 [3.2K]

Answer:

x = \dfrac{7 \pm 3i\sqrt{3}}{2}

Step-by-step explanation:

(x – 5)^2 + 3(x – 5) + 9 = 0

This is a quadratic equation in x - 5.

Let u = x - 5, then the quadratic equation becomes:

u^2 + 3u + 9 = 0

We can use the quadratic formula to solve for u.

u = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

u = \dfrac{-3 \pm \sqrt{3^2 - 4(1)(9)}}{2(1)}

u = \dfrac{-3 \pm \sqrt{9 - 36}}{2}

u = \dfrac{-3 \pm \sqrt{-27}}{2}

u = \dfrac{-3 \pm 3i\sqrt{3}}{2}

Since u = x - 5, now we substitute x - 5 for u and solve for x.

x - 5 = \dfrac{-3 \pm 3i\sqrt{3}}{2}

x = \dfrac{-3 \pm 3i\sqrt{3}}{2} + 5

x = \dfrac{-3 \pm 3i\sqrt{3}}{2} + \dfrac{10}{2}

x = \dfrac{7 \pm 3i\sqrt{3}}{2}

7 0
3 years ago
A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represe
Stella [2.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

            P(D_{1} ) = 0.12                                   P(D_{2} ) = 0.07

           P(D_{3} ) = 0.05                                    P (D_{1} U D_{2} ) = 0.13

          P(D_{1}n D_{2}n D_{3}) = 0.01                        P(D_{1} U D_{3}) = 0.14

Step 2 : Obtain the probability that a unit does not have a type 1 defect

         P(\frac{}{D_{1} }) =  1 -P(D_{1} )

                    = 1 - 0.12

                    = 0.88  

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

          The probability of the unit having both type 2 and type 3 defect is denoted as P(D_{2} n D_{3} )

   This is calculated as

                    P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\                          = 0.07 + 0,05 - 0.13

                    =   0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

                  Let P(\frac{}{D_{1}} n D_{2} n D_{3} ) denote the  probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

                      P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})

                                               =   0.02 - 0.01

                                               =  0.01

Step 4 : Obtain the probability that a unit has at most two defects

               P(at most 2 defects)  = 1 - P(all three defects)

                                                  = 1- P(D_{1} n D_{2}nD_{3})

                                                  =  1 - 0.01

                                                  = 0.99

7 0
3 years ago
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