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Ad libitum [116K]
2 years ago
15

Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula?

Quadratic formula: x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (3)(6) EndRoot Over 2(3) EndFraction x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4 (3)(6) EndRoot Over 2(3) EndFraction
Mathematics
2 answers:
Stella [2.4K]2 years ago
4 0

Answer:

x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

0=-3x^{2}-2x+6  

so

a=-3\\b=-2\\c=6

substitute in the formula

x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(-3)(6)}} {2(-3)}

therefore

x = StartFraction negative (negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction

Rom4ik [11]2 years ago
3 0

Answer:

its a

Step-by-step explanation:

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
2 years ago
Noor brought 212121 sheets of stickers. She gave \dfrac{1}{3} 3 1 ​ start fraction, 1, divided by, 3, end fraction of a sheet to
My name is Ann [436]

Answer:

12 teachers will get 1/2 sheets each

Step-by-step explanation:

Noor bought 21 sheets of stickers

She gave 1/3 sheets to each 45 students

She wants to give teachers 1/2 of sheets

The question should be: how many teachers will she give

Total sheets=21

Students gets= 1/3 × 45

=45/3

=15 sheets

Remaining sheet= total sheets - students sheets

=21-15

=6 sheets

There are 6 sheets remaining for teachers

She wants to give 1/2 to each teacher

Then,

The number of teachers that will get =Remaining sheets ÷ each teacher's share

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=6 × 2/1

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12 teachers will get 1/2 sheets each

5 0
2 years ago
Points H and F lie on circle c What is the length of line segment GH?
Nina [5.8K]

Answer:

6 units

Step-by-step explanation:

Given: Points H and F lie on  circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.

To find: Length of GH.

Sol: EC = CH = 9 (Radius of the same circle are equal)

Now, GC = GH + CH

GC = GH + 9

Now In ΔEGC, using pythagoras theorem,

(Hypotenuse)^{2} = (Base)^{2} +(Altitude)^{2} ......(ΔEGC is a right triangle)

(GC)^{2} = (GE)^{2} +(EC)^{2}

(GH + 9)^{2} = (9)^{2} +(12)^{2}

(GH )^{2} + (9)^{2} + 18GH = 81 + 144

(GH )^{2} + 81 + 18GH = 81 + 144

(GH )^{2} + 18GH = 144

Now, Let GH = <em>x</em>

x^{2} +18x = 144

On rearranging,

x^{2} +18 x - 144 = 0

x^{2} - 6x +24x + 144 = 0

x (x-6) + 24 (x - 6) =0

(x - 6) (x + 24) = 0

So x = 6  and x = - 24

∵ x cannot be - 24 as it will not satisfy the property of right triangle.

Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.

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2 years ago
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She would be able to download 49 at this rate.
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2 years ago
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Suppose that 40 batteries are shipped to an auto parts store, and that 4 of those are defective. A fleet manager then buys 8 of
Assoli18 [71]
From the given above, we will include the case in which 3 and all 4 defectives are included in the purchase. 

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2 years ago
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