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dsp73
3 years ago
10

This is due today, someone pls help!

Mathematics
2 answers:
Alex3 years ago
6 0
The answer should be 32,890!

Move the decimal 4 places to the right
3.289 —> 32890

Jesse moved it 7 places to the right!

I’m not great at explaining but I hope it’ll help!
Juli2301 [7.4K]3 years ago
3 0
The answer to this should be 32,890
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Two sets of equatic expressions are shown below in various forms: Line 1: x2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)2 − 0.25 Line 2: x
kherson [118]

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.

7 0
3 years ago
Determine y when x = 2.5, if y = 144 when x = 8.
Mrac [35]

Answer:

=144*8/2.5

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Step-by-step explanation:

there you go

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Why is the polynomial, 4x^2y + 5xy classified as a 3rd degree binomial?
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