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3 years ago

Choose the constant term that completes the perfect square trinomial.

1 answer:

3 0

A perfect trinomial, if we begin by a binomial is defined as: the square of the first term, plus (or minus) the double product of the first term times the second, plus the square of the second term:
(a + b)^2 = a^2 + 2ab + b^2
We are given:
y^2 + 5y + x
we need to find x, so x is defined as a squared quantity, which is equal to the second term coefficient (5) divided by 2, and that number squared, that is:
(5/2)^2 = 25/4
that is the third term for the trinomial to be perfect.

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mihalych1998 [28]

Answer:

y = ⅓x

Step-by-step explanation:

Let's find m.

m = y/x

Using the point on the line, (3, 1),

m = ⅓

Substitute m = ⅓ into y = mx

This, the equation that describes the relationship would be:

y = ⅓x

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3 years ago

Does anyone know how to do this help please

statuscvo [17]

Answer:

$\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}$

Step-by-step explanation:

A quadratic function is given to us . And we need to fill out the table by using the function . The given function is ,

$\rm\implies y = x^2+x - 6$

Here we need to substitute the different values of x , to get the different values of y.

Put x = -3 :-

$\rm\implies y = 3^2-3-6\\$

$\rm\implies y = 9 - 3 - 6 \\$

$\rm\implies y = 0$

Put x = -2 :-

$\rm\implies y = 2^2-2-6\\$

$\rm\implies y = 4 -2-6\\$

$\rm\implies y = -4$

Put x = -1 :-

$\rm\implies y = -1^2-1-6\\$

$\rm\implies y = 1 -1-6 \\$

$\rm\implies y = -6$

Put x = 1 :-

$\rm\implies y = 1^2+1-6\\$

$\rm\implies y = 2 -6$

$\rm\implies y = -4$

Put x = 2 :-

$\rm\implies y = 2^2+2-6\\$

$\rm\implies y = 4 +2-6\\$

$\rm\implies y = 2$

Put x = 3 :-

$\rm\implies y = 3^2+3-6\\$

$\rm\implies y = 9 +3-6\\$

$\rm\implies y = 6$

Final table :-

$\boxed{\begin{array}{c|c|c|c|c|c|c|c} \bf x &\rm -3 &\rm -2 &\rm -1 &\rm 0 &\rm 1 &\rm 2 &\rm 3 \\\\ \bf y & \rm 0 &\rm -4 &\rm -6&\rm -6 &\rm -4&\rm 2 &\rm 6 \end{array}}$

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