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Rzqust [24]
2 years ago
5

{x + y = 2

Mathematics
2 answers:
Mila [183]2 years ago
7 0

1) I put them in two separate brakets.

2) I solved it by equating both of them.

Studentka2010 [4]2 years ago
4 0

Answer:

  1. Substitute for y in the first equation. Solve for x by completing the square. Substitute for x in the first equation to find y.

  2. (x, y) = (2+2√2, -2√2) and (2-2√2, 2√2)

Step-by-step explanation:

<h3>1.</h3>

The first equation is a linear equation in standard form. The second equation is a quadratic equation giving an expression for y. The system can be solved by using the second equation to substitute for y in the first equation. After the resulting equation is put in suitable form, the irrational solutions can be found by completing the square. Then the y-values can be found by substituting the x-values into either of the given equations. Usually, it is easiest to substitute into the linear equation.

__

<h3>2.</h3>

Substituting for y in the first equation:

  x +(-1/4x^2 +3) = 2

Multiplying by -4 to eliminate the fraction and make the leading coefficient 1, we have ...

  -4x +x^2 -12 = -8

  x^2 -4x = 4 . . . . . . add 12

  x^2 -4x +4 = 8 . . . . add 4 to complete the square

  (x -2)^2 = 8 . . . . . . write as a square

  x -2 = ±√8 = ±2√2 . . . . take the square root

  x = 2 ±2√2 . . . . . . add 2

Solving the first equation for y, we get ...

  y = 2 -x

Substituting the values of x we found gives ...

  y = 2 -(2 ±2√2) = ∓2√2

Solutions are ...

  (x, y) = (2+2√2, -2√2) and (2-2√2, 2√2)

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A construction firm bids on two different contracts. Let E1 be the event that the bid on the first contract is successful, and d
marissa [1.9K]

Calculate the probability that both bids are successful

Answer:

The probability that both contracs are successful is 0.21

Step-by-step explanation:

Given

E1 = the event that the bid on the first contract is successful

E2 = the event that the bid on the second contract is successful

P(E1) = 0.3

P(E2) = 0.7

Let P(A) represent the event that both contracts are successful

P(A) = P(E1 and E2)

Since both events are independent. P(A) becomes

P(A) = = P(E1 * P(E2)

By substituton

P(A) = 0.3 * 0.7

P(A) = 0.21

Hence the probability that both contracs are successful is 0.21

7 0
3 years ago
Translate into algebraic expression and simplify: ''Ten times the sum of a number and three'' b) nine substracted from twice a n
hichkok12 [17]

Step-by-step explanation:

a)\\n-\text{the number}\\\\\text{Ten times the sum of a number and three:}\\\\\boxed{10\times(n+3)=10(n+3)}\\\\b)\\n-\text{the number}\\\\\text{Nine substracted from twice a number}:\\\\\boxed{2n-9}

4 0
3 years ago
The ratio of the surface areas of two similar solids is 49:100. What is the ratio of their corresponding side lengths?
algol13
<span>ratio of surface area = 49:100
ratio of corresponding lengths = √49:√100 = 7:10
answer: C

Hope this helps


</span>
3 0
3 years ago
Read 2 more answers
23n=2.4<br> A. 3.6 <br><br> B. 7.2 <br><br> C. 1.2<br><br> D. 1.6
svp [43]

\huge\text{Hey there!}

\mathsf{23n =2.4}

\large\text{DIVIDE 23 to BOTH SIDES}

\mathsf{\dfrac{23n}{23}=\dfrac{2.4}{23}}

\large\text{Cancel out: }\mathsf{\dfrac{23}{23}}\large\text{ because it gives you 1}

\large\text{Keep: }\mathsf{\dfrac{2.4}{23}}\large\text{ because it helps us solve for n}

\mathsf{n = \dfrac{2.4}{23}}

\mathsd{\dfrac{2.4}{23}= 2.4\div 23\rightarrow \bf 0.104348}

\boxed{\boxed{\large\textsf{Answer: \huge n = \bf 0.104348}}}\huge\checkmark

\boxed{\textsf{NONE OF THE ABOVE is your ANSWER}}

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

8 0
2 years ago
A movie theater has a seating capacity of 329. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.
kipiarov [429]

Answer:

170 children

74 students

85 adults

Step-by-step explanation:

Given

Let:

C = Children; S = Students; A = Adults

For the capacity, we have:

C + S + A = 329

For the tickets sold, we have:

5C + 7S + 12A = 2388

Half as many as adults are children implies that:

A = \frac{C}{2}

Required

Solve for A, C and S

The equations to solve are:

C + S + A = 329 -- (1)

5C + 7S + 12A = 2388 -- (2)

A = \frac{C}{2} -- (3)

Make C the subject in (3)

C = 2A

Substitute C = 2A in (1) and (2)

C + S + A = 329 -- (1)

2A + S + A = 329

3A + S = 329

Make S the subject

S = 329 - 3A

5C + 7S + 12A = 2388 -- (2)

5*2A + 7S + 12A = 2388

10A + 7S + 12A = 2388

7S + 22A = 2388

Substitute S = 329 - 3A

7(329 - 3A) + 22A = 2388

2303 - 21A + 22A = 2388

2303 +A = 2388

Solve for A

A = 2388 - 2303

A = 85

Recall that: C = 2A

C = 2 * 85

C = 170

Recall that: S = 329 - 3A

S = 329 - 3 * 85

S = 329 - 255

S = 74

Hence, the result is:

C = 170

S = 74

A = 85

8 0
3 years ago
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