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3 years ago

When does an absolute value equation have no solution? Help pls

2 answers:

5 0

The first thing that comes to mind that an absolute value cannot have a nagative output, in other words, if you have an function that is equal to a negative number, then there is no solution.

mixer [17]3 years ago

3 0

Hii! I'm nat and I'm gonna help ya!

An absolute value expression has absolutely no solution if the absolute value expression equals a negative number!

Hope this helps !

Brainliest?

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Express 0.456 as a percent

nordsb [41]

<span>Express 0.456 as a percent
=
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3 years ago

Read 2 more answers

(25+(2×7)+3×(6-3)+2000

Harrizon [31]

Answer: 2048

Step-by-step explanation:

(25+(2×7)+3×(6-3)+2000

25+14+(3x3)+2000
39+9+2000
48+2000
2048

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2 years ago

I require assistance please.

Zolol [24]

The answer is b because all of the other numbers in the parentheses are less then one so there can’t be a growth if the number is less than one

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3 years ago

Need help ASAP please<br> How many solutions are there to the following equation:<br> a+5 = -5+5

Aliun [14]

Answer:

1 solution

a=-5

Step-by-step explanation:

4 0

3 years ago

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Which table contains the vertex of the graph represented by y=x^2-6x+4?

notka56 [123]

Answer:

Table B

Vertex is (3,-5)

Step-by-step explanation:

We are given with an equation of a parabola $y=x^2-6x+4$

Let is convert it into standard form of a parabola

$y=x^2-6x+4$

adding and subtracting 9 in the right hand side of the =

$y=x^2-6x+9-9+4$

$y=x^2-2\times 3\times x+ 3^2-9+4$

the first three terms of the right hand side forms the expression of square of difference

$a^2-2 \times a \times b+b^2 = (a-b)^2$

Hence

$y=(x-3)^2-5$

adding 5 on both sides we get

$(y+5)=(x-3)^2$

Comparing it with the standard equation of a parabola

$X^2=4\times \frac{1}{4} \times Y$

where $X=x-3$ and $Y=y+5$

The vertex of $X^2=4\times \frac{1}{4} \times Y$ will be (0,0)

and thus vertex of

$(y+5)=(x-3)^2$ will be (3,-5)

Hence the Table B is our right answer

5 0

3 years ago