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Nutka1998 [239]
2 years ago
14

The question is the image

Mathematics
1 answer:
oksian1 [2.3K]2 years ago
7 0

Answer:

(- 4, 16 ) , (\frac{1}{2}, \frac{1}{4} )

Step-by-step explanation:

y = x² → (1)

7x + 2y = 4 → (2)

substitute y = x² into (2)

7x + 2x² = 4 ( subtract 4 from both sides )

2x² + 7x - 4 = 0 ← in standard form

(x + 4)(2x - 1) = 0 ← in factored form

equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = \frac{1}{2}

substitute these values into (1) for corresponding values of y

x = - 4 : y = (- 4)² = 16 ⇒ (- 4, 16 )

x = \frac{1}{2} : y = (\frac{1}{2} )² = \frac{1}{4} ⇒ ( \frac{1}{2} , \frac{1}{4} )

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Leroy took a trip to Alaska. His airfare was $243 and he spent $72 each day while
Aleks04 [339]

Answer:

Total cost of the trip = 243 + 72x

Step-by-step explanation:

Cost of airfare = $243

Amount spent per day = $72

Which expression can he use to help him find the cost of the trip?

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x = number of days spent in Alaska

Total cost of the trip = Cost of airfare + (Amount spent per day * number of days spent in Alaska)

= 243 + (72 * x)

= 243 + 72x

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Total cost of the trip = 243 + 72x

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2 years ago
Caitlin has $5 and $10 bills that are worth $675. She has twice as many $10 bills as $5 bills. How many of each type of bill doe
Lerok [7]

Caitlin has 27 $5 bills and 54 $10 bills.

Step-by-step explanation:

Given,

Worth of $5 and $10 bills = $675

Let,

Number of $5 bills = x

Number of $10 bills = y

According to given statement;

5x+10y=675    Eqn 1

She has twice as many $10 bills as $5 bills.

y=2x    Eqn 2

Putting value of y from Eqn 2 in Eqn 1

5x+10(2x)=675\\5x+20x=675\\25x=675

Dividing both sides by 25

\frac{25x}{25}=\frac{675}{25}\\x=27

Putting x=27 in Eqn 2

y=2(27)\\y=54

Caitlin has 27 $5 bills and 54 $10 bills.

Keywords: linear equation, substitution method

Learn more about substitution method at:

  • brainly.com/question/10699220
  • brainly.com/question/10703930

#LearnwithBrainly

7 0
3 years ago
A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the eve
tester [92]

Answer:

The indicated probability of P(D \cup F')=\frac{25}{26}

Step-by-step explanation:

Probability of an event E to be;

P(E) = \frac{Number of events within E}{Total number of possible outcomes}

As per the given condition:

Total number of possible outcomes =  52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.    

Then,

From the given condition:

P(D) = \frac{26}{52}   [Out of 52 cards, 26 were black] ,

P(F) = \frac{4}{52}    [Out of 52 cards, there are four  10 cards]

For any two events A and B we always have;

P(A \cup B) = P(A)+P(B)-P(A \cap B)

Now, we have to find the indicated probability:

P(D \cup F')=P(D)+P(F')-P(D \cap F')               ......[1]

First find the P(F');

P(F') =1-P(F) = 1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}

Also, to find P(D \cap F').

We use the formula :

For any event A and B independent variable.

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then;

P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}

Now, substitute these in [1];

P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}=\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}

Therefore, the probability of P(D \cup F')=\frac{25}{26}



3 0
3 years ago
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