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algol13
3 years ago
6

Patel is solving 8x2 + 16x + 3 = 0. Which steps could he use to solve the quadratic equation? Select three options.

Mathematics
2 answers:
MakcuM [25]3 years ago
7 0

Answer:

<em>8(x2 + 2x) = –3 </em>

<em>8(x2 + 2x + 1) = –3 + 8 </em>

<em>x = –1 Plus or minus StartRoot StartFraction 5 Over 8 EndFraction EndRoot</em>

Step-by-step explanation:

<u>Solving Quadratic Equations </u>

Sometimes it's preferred to solve quadratic equations without the use of the known quadratic formula solver. One of the most-used methods consists of completing squares and solving for x.

We have the equation

\displaystyle 8x^2+16x+3=0

We separate variables from constants

\displaystyle 8x^2+16x=-3

Taking the common factor  8

\displaystyle 8(x^2+2x)=-3

Completing squares in the brackets and balancing the equation in the right side

\displaystyle 8(x^2+2x+1)=-3+8

Factoring the perfect square

\displaystyle 8(x+1)^2=5

Isolating x

\displaystyle (x+1)^2=\frac{5}{8}

\displaystyle (x+1)=\pm \sqrt{\frac{5}{8}}

\displaystyle x=-1\pm \sqrt{\frac{5}{8}}

We can clearly see the steps used to solve the quadratic equation are (in order and written like in the question)

8(x2 + 2x) = –3

8(x2 + 2x + 1) = –3 + 8

x = –1 Plus or minus StartRoot StartFraction 5 Over 8 EndFraction EndRoot

marshall27 [118]3 years ago
3 0

Answer:

A B D

Step-by-step explanation:

I got a 100 on edge

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Amy scored 13,281 points. 
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Determine the equation of each line
katovenus [111]

Answer:

see explanation

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

(a)

here m = - \frac{3}{4} and c = 6, hence

y = - \frac{3}{4} x + 6 ← equation of line

(b)

here m = 6, hence

y = 6x + c ← is the partial equation

to find c substitute (2, - 6 ) into the partial equation

- 6 = 12 + c ⇒ c = - 6 - 12 = - 18

y = 6x - 18 ← equation of line

(c)

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (- 1, 3) and (x₂, y₂ ) = (4, 7)

m = \frac{7-3}{4+1} = \frac{4}{5}, hence

y = \frac{4}{5} x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (- 1, 3 ), then

3 = - \frac{4}{5} + c → c = 3 + \frac{4}{5} = \frac{19}{5}

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3 years ago
On average 83.1% of welds performed by a particular welder are defective. if, in one day, the welder creates three welds. what i
Nuetrik [128]
Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.

Using the binomial distribution, we want all three to be defective.
P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}
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3 years ago
A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev
Julli [10]

Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

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2 years ago
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olganol [36]

\frac{1}{5}x -  \frac{2}{3} y = 30  \\

Putting the value of Y = 15

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So, X = 200

6 0
2 years ago
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