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3 years ago

|x| > - 16 what is the value of x

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

8 0

9514 1404 393

Answer:

x ∈ all real numbers

Step-by-step explanation:

The given inequality is true for any value of x. The absolute value of a quantity is always greater than any negative number.

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Which number is irrational? A. Syntax error from line 1 column 53 to line 1 column 60. Unexpected '0'. B. 1 over 9 C. square roo

DerKrebs [107]

C. because the decimal is never ending, so it is irrational.

8 0

3 years ago

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Expand the binomial<br> <img src="https://tex.z-dn.net/?f=%28x-%5Cfrac%7B2%7D%7B5%7D%29%5E%7B2%7D" id="TexFormula1" title="(x-\f

kkurt [141]

Answer:

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

Step-by-step explanation:

You have two methods to expand this binomial.

Method 1

If you have the expression:

$(x- \frac{2}{5})^{2}$

You can write the expression it in the following way:

$(x-\frac{2}{5})^{2}=(x-\frac{2}{5})(x-\frac{2}{5})$

Then, apply the distributive property:

$(x-\frac{2}{5})(x-\frac{2}{5}) = x^2 -\frac{2}{5}x -\frac{2}{5}x+ (\frac{2}{5})\frac{2}{5}$

Simplify the expression:

$(x-\frac{2}{5})^2= x^2 -\frac{4}{5}x+ (\frac{4}{25})$

---------------------------------------------------------------------------------------

Method 2

For any expression of the form:

$(a-b)^2$

Its expanded form will be:

$(a-b)^2= a^2 -2ab + b^2$

$a = x$

$b =\frac{2}{5}$

$(x- \frac{2}{5})^{2} = x^2 - 2x\frac{2}{5} + (\frac{2}{5})^2$

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

3 0

3 years ago

Describe and correct the error in determining wheather (8,11) is a solution of y-x=3

Burka [1]

(8, 11) is the correct solution. x = 8, y = 11

Plug it in:

11-8=3

That's correct.

7 0

3 years ago

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What is the benefit of dividing out greatest common factors (GCFs) before multiplying?

Nina [5.8K]

Lets get an example, for this example lets use 24/30 x 30/60
Without dividing out GCFs you would get 24x12 over 30 x 60 which is 288/1800
With dividing out GCFs you get 2/1 x 5/1 Which is 7 :)

7 0

3 years ago

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Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago