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Gennadij [26K]
3 years ago
10

5. Simplify (7x2+ 8) - (x2 - 2)

Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

(7 \times 2 + 8) - ( {x}^{2}  - 2) \\ (14 + 8) - ( {x}^{2}  - 2) \\ 14 + 8 -  {x}^{2}  + 2 \\ 22 +  {x}^{2}  + 2 \\ 24 +  {x}^{2}

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

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Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22. (Past studies have revealed that onl
Alex17521 [72]

Answer:

B. no, it is not low enough

A. no, it is not low enough

Step-by-step explanation:

Given that Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22.

Prob for  a random person booked arrive for flight = 0.86

No of persons who books and arrive for flight, X is binomial, since there are two outcomes and each person is independent of the other

The probability that if Air-USA books 24 persons, not enough seats will be available

= P(X=23)+P(x=24)

= 0.1315

B. no, it is not low enough

-------------------------------

The prob we got is >10% also

A. no, it is not low enough

7 0
3 years ago
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
Subtract (x^2+3x-4) -(4x-2x^2-3)
Feliz [49]

Answer:

3 x^2 - x + -1

Step-by-step explanation:

Simplify the following:

-(4 x - 2 x^2 - 3) + x^2 + 3 x - 4

Factor -1 out of -2 x^2 + 4 x - 3:

--(2 x^2 - 4 x + 3) + x^2 + 3 x - 4

(-1)^2 = 1:

2 x^2 - 4 x + 3 + x^2 + 3 x - 4

Grouping like terms, 2 x^2 + x^2 + 3 x - 4 x - 4 + 3 = (x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3):

(x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3)

x^2 + 2 x^2 = 3 x^2:

3 x^2 + (3 x - 4 x) + (-4 + 3)

3 x - 4 x = -x:

3 x^2 + -x + (-4 + 3)

3 - 4 = -1:

Answer: 3 x^2 - x + -1

4 0
4 years ago
Figure ABCDE was reflected across the line y = r to
marysya [2.9K]

Answer:

-2.6

Step-by-step explanation:

7 0
3 years ago
What value of n makes the equation true? (2x^9y^n)(4x^2y^10)=8x^11y^20
Komok [63]
Hi there! The answer is n = 10.

(2 {x}^{9}  {y}^{n} )(4 {x}^{2}  {y}^{10} ) = 8 {x}^{11}  {y}^{20}

As you see at the powers of x, we need to add the exponents of the power we when multiply them.
{x}^{9}  \times  {x}^{2}  =  {x}^{9 + 2}  = x {}^{11}

The powers of y work the same way.
{y}^{n}  \times  {y}^{10}  =  {y}^{n + 10}  =  {y}^{20}

Hence, n = 10, since
{y}^{10 + 10}  =  {y}^{20}
8 0
3 years ago
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