2 years ago

What is the square root of ten

Mathematics

1 answer:

lyudmila [28]2 years ago

8 0

10 is not a perfect square I believe

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Find the inverse for each relation. 1. { (1, -3), (-2, 3), (5, 1), (6, 4) }

velikii [3]

You just flip each of the numbers spots - the x and y

3 0

2 years ago

Find the values of x when y=1

kirill [66]

Answer:

$x=1+\sqrt{5}, 1-\sqrt{5}$

Step-by-step explanation:

$x^{2} -2x-4=0$
$(x-1)^{2} -1-4=0$
$(x-1)^{2} -5=0$
$(x-1)^{2} =5$
$x-1=\sqrt{5} ,-\sqrt{5}$
$x=1+\sqrt{5} , 1-\sqrt{5}$

FINAL ANSWER

$x=1+\sqrt{5} , 1-\sqrt{5}$

In Decimal Form (Rounded to 3 significant figures)

x≈-1.24, 3.24

6 0

3 years ago

Read 2 more answers

How do you simplify this

Dafna11 [192]

Top:

x / (x + 1) - 1 / x

= [x^2 - (x +1)] / x(x+1)

= (x^2 - x - 1 ) / x (x+1)

Bottom:

x / (x + 1) + 1 / x

= [x^2 + (x +1)] / x(x+1)

= (x^2 + x + 1 ) / x (x+1)

Now you have:

(x^2 - x - 1 ) / x (x+1)

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

(x^2 + x + 1 ) / x (x+1)

= (x^2 - x - 1 ) / x (x+1) * x (x+1) / (x^2 + x + 1 )

= (x^2 - x - 1 ) /(x^2 + x + 1 )

Answer:

x^2 - x - 1

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

x^2 + x + 1

7 0

2 years ago

I cannot solve this.. i want to solve this

GuDViN [60]

Can’t see that properly can’t you write it bigger or something

3 0

2 years ago

PLZZ HELP!

dezoksy [38]

For the first one it’s h=6 and the second one is h=-3

4 0

3 years ago