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geniusboy [140]
3 years ago
15

Simplify completely 4x^2-7x+3 / x^2 + 5x-6

Mathematics
2 answers:
babunello [35]3 years ago
5 0

Answer:

Your answer is D.

\frac{4x-3}{x+6}

Step-by-step explanation:

First, you have to factor by grouping.

Next, factor using AC method.

Last, cancel the common factor.

Now all you do is rewrite the expression.






krek1111 [17]3 years ago
4 0

Answer:

D)   \frac{4x-3} {x+6}

Step-by-step explanation:

First we have to factorize and then cancel the like terms.

Given,

\frac{4x^2-7x+3} {x^{2}+5x-6}\\=\frac{4x^2-3x-4x+3} {x^{2}+6x-x-6}\\=\frac{x(4x-3)-1(4x-3)} {x(x+6)-1(x+6)}\\=\frac{(4x-3)(x-1)} {(x+6)(x-1)}

Now, we got the like term  (x-1).

Canceling this like term we get

\frac{(4x-3)(x-1)} {(x+6)(x-1)}=\frac{4x-3} {x+6}

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Step-by-step explanation:

1/2 = 4/8

=> 1/8 < 4/8

=> 1/8 < 1/2

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A sports club offers an organized basketball league. A team pays $600 to join their league. In addition to paying their share of
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5

Step-by-step explanation:

600-775=125

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Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Vis
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Answer:

Step-by-step explanation:

Given that,

Visa card is represented by P(A)

MasterCard is represented by P(B)

P(A)= 0.6

P(A')=0.4

P(B)=0.5

P(B')=0.5

P(A∩B)=0.35

1. P(A U B) =?

P(A U B)= P(A)+P(B)-P(A ∩ B)

P(A U B)=0.6+0.5-0.35

P(A U B)= 0.75

The probability of student that has least one of the cards is 0.75

2. Probability of the neither of the student have the card is given as

P(A U B)'=1-P(A U B)

P(A U B)= 1-0.75

P(A U B)= 0.25

3. Probability of Visa card only,

P(A)= 0.6

P(A) only means students who has visa card but not MasterCard.

P(A) only= P(A) - P(A ∩ B)

P(A) only=0.6-0.35

P(A) only=0.25.

4. Compute the following

a. A ∩ B'

b. A ∪ B'

c. A' ∪ B'

d. A' ∩ B'

e. A' ∩ B

a. A ∩ B'

P(A∩ B') implies that the probability of A without B i.e probability of A only and it has been obtain in question 3.

P(A ∩ B')= P(A-B)=P(A)-P(A∩ B)

P(A∩ B')= 0.6-0.35

P(A∩ B')= 0.25

b. P(A ∪ B')

P(A ∪ B')= P(A)+P(B')-P(A ∩ B')

P(A ∪ B')= 0.6+0.5-0.25

P(A ∪ B')= 0.85

c. P(A' ∪ B')= P(A')+P(B')-P(A' ∩ B')

But using Demorgan theorem

P(A∩B)'=P(A' ∪ B')

P(A∩B)'=1-P(A∩B)

P(A∩B)'=1-0.35

P(A∩B)'=0.65

Then, P(A∩B)'=P(A' ∪ B')= 0.65

d. P( A' ∩ B' )

Using demorgan theorem

P(A U B)'= P(A' ∩ B')

P(A U B)'= 1-P(A U B)

P(A' ∩ B')= 1-0.75

P(A' ∩ B')= 0.25

P(A U B)'= P(A' ∩ B')=0.25

e. P(A' ∩ B)= P(B ∩ A') commutative law

Then, P(B ∩ A') = P(B) only

P(B ∩ A') = P(B) -P(A ∩ B)

P(B ∩ A') =0.5 -0.35

P(B ∩ A') =0.15

P(A' ∩ B)= P(B ∩ A') =0.15

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Answer: The submarine would be 710 feet below the surface after it is finished diving.

Step-by-step explanation:

Given: Each move downward was of 355 feet

If the submarine dives below the surface, heading downward in two moves.

Then, the total distance moved by submarine = 2 (355) feet

= 710 feet

Therefore, the submarine would be 710 feet below the surface after it is finished diving.

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