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Firlakuza [10]
3 years ago
7

\begin{aligned} &-5y=-5 \\\\ &7x+6y=7 \end{aligned}

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Answer:

(5,1) is not a solution of the above system of equations.

Step-by-step explanation:

We are given a system of equations that are  

- 5y = - 5 ......... (1) and  

7x + 6y = 7 ........... (2)

So, solving equation (1) we get y = 1 and putting y = 1 in the equation (2) we get 7x = 7 - 6(1) = 1

⇒ x = \frac{1}{7}

So, it clear from the solution of the above equations i.e. (\frac{1}{7}, 1), we can say that (5,1) is not a solution of the above system of equations. (Answer)

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The table shows some values of a function of the form y = ax2 + bx + c.
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Answer:

Value of constant term c is (-4)

Step-by-step explanation:

The given table represents a function which is in the form of a quadratic equation,

y = ax² + bx + c

We choose three points (3, -10), (4, -16) and (5, -24) from the table and satisfy the equation to get the values of a, b, and c.

For point (3, -10)

-10 = a(3)² + 3b + c

9a + 3b + c = -10 -------(1)

For point (4, -16)

-16 = a(4)² + 4b + c

16a + 4b + c = -16 ------(2)

For point (5, -24)

-24 = a(5)² + 5b + c

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Equation (1) - equation (2)

(9a + 3b + c) - (16a + 4b + c) = -10 + 16

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7a + b = -6 ------(4)

Equation (2) - equation (3)

(16a + 4b + c) - (25a + 5b + c) = -16 + 24

-9a - b = 8

9a + b = -8 -------(5)

Equation (4) - Equation (5)

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From equation (4),

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b = 1

From equation (1)

9(-1) + 3(1) + c = -10

-9 + 3 + c = -10

c = -10 + 6

c = -4

Therefore, the value of constant term c is (-4).

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