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Dmitry [639]
2 years ago
14

Find the domain of the function f of x equals the cube root of the quantity x plus three, minus one.

Mathematics
2 answers:
kifflom [539]2 years ago
8 0
<span>The domain of the function f of x equals the cube root of the quantity x plus three, minus one would be:

ALL REAL NUMBERS 

I hope my answer has come to your help. God bless and have a nice day ahead!


</span>
Fantom [35]2 years ago
8 0
Hello there.

<span>Find the domain of the function f of x equals the cube root of the quantity x plus three, minus one.

All real Numbers </span>
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Nesterboy [21]

Answer:

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Step-by-step explanation:

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2 years ago
Inverse of g(x) 2x+3
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2 years ago
Can I please have help with question 2) a and b
Ulleksa [173]

Answer:

\frac{10\sqrt{6} }{3} and \frac{11\sqrt{3} }{3}

Step-by-step explanation:

(a)

A = \frac{1}{2} × 4\sqrt{2} × \frac{5}{\sqrt{3} }

   = 2\sqrt{2} × \frac{5}{\sqrt{3} }

   = \frac{10\sqrt{2} }{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

= \frac{10\sqrt{6} }{3} units²

(b)

Using Pythagoras' identity in the right triangle

let hypotenuse be h , then

h² = (4\sqrt{2} )² + (\frac{5}{\sqrt{3} } )²

    = 32 + \frac{25}{3}

    = \frac{96}{3} + \frac{25}{3}

    = \frac{121}{3} ( take the square root of both sides )

h = \sqrt{\frac{121}{3} } = \frac{11}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

h = \frac{11\sqrt{3} }{3}

 

7 0
3 years ago
8 3/4 + 9 5/9= <br> with working?
Gre4nikov [31]

8 3/4 + 9 5/9 Add 8 and 9

17 + 3*9/4*9 + 5*4/9*4 Find common denominator

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If you need the decimal form, it is 18.31. But other than that, the answer is 18 11/36.

4 0
3 years ago
Help Please! Giving 98 points!
shusha [124]

Answer:

the anser is B. btw you should try the website Desmos. Its a graph calculater

Step-by-step explanation:


4 0
3 years ago
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