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3 years ago

Evaluate rs+14s when R=6 S=1/4

Mathematics

2 answers:

dezoksy [38]3 years ago

8 0

Never mind all that. I messed up

krok68 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

plug in the values for r and s and solve.

(6)(1/4)+14(1/4)

(3/2)+(7/2)

(10/2)

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Answer: The correct line is

$\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.$

Step-by-step explanation: We are given the following two sets of quadratic expressions in various forms:

$\textup{Line 1: }x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25,\\\\\textup{Line 2 :}x^2+5x+6=(x+2)(x+3)=(x+2.5)^2+6.25.$

We are to select one of the lines from above that represent three equivalent expressions.

We can see that there are three different forms of a quadratic expression in each of the lines:

First one is the simplified form, second is the factorised form and third one is the vertex form.

So, to check which line is correct, we need to calculate the factorised form and the vertex form from the simplified form.

We have

$\textup{Line 1: }\\\\x^2+3x+2\\\\=x^2+2x+x+2\\\\=x(x+2)+1(x+2)\\\\=(x+1)(x+2),$

and

$x^2+3x+2\\\\=x^2+2\times x\times 1.5+(1.5)^2-(1.5)^2+2\\\\=(x+1.5)^2-2.25+2\\\\=(x+1.5)^2-0.25.$

So,

$\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.$

Thus, Line 1 contains three equivalent expressions.

Now,

$\textup{Line 2: }\\\\x^2+5x+6\\\\=x^2+3x+2x+6\\\\=x(x+3)+2(x+3)\\\\=(x+2)(x+3),$

and

$x^2+5x+6\\\\=x^2+2\times x\times 2.5+(2.5)^2-(2.5)^2+6\\\\=(x+2.5)^2-6.25+6\\\\=(x+2.5)^2-0.25\neq (x+2.5)^2+6.25.$

So,

$\textup{Line 2 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2+6.25.$

Thus, Line 2 does not contain three equivalent expressions.

Hence, Line 1 is correct.

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