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andreyandreev [35.5K]
2 years ago
11

Rationalise 1/√7-√3PLEASE GIVE ANSWER IT'S URGENT ​

Mathematics
1 answer:
IgorLugansk [536]2 years ago
5 0

\dfrac 1{\sqrt  7- \sqrt 3}\\\\\\=\dfrac{\sqrt 7 + \sqrt 3}{\left ( \sqrt 7 -\sqrt 3\right) \left( \sqrt 7 + \sqrt 3\right)}\\\\\\=\dfrac{\sqrt 7 + \sqrt 3}{\left(\sqrt  7 \right)^2 - \left(\sqrt 3 \right)^2}\\ \\\\=\dfrac{\sqrt 7 + \sqrt 3}{7-3}\\\\\\=\dfrac  14\left(\sqrt 7 + \sqrt 3\right)

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there are 8 students on the minibus. five of the students are boys. what fraction of the students are boy
meriva
Students on the minibus = 8
students that are boys = 5
fraction of the student that are boys = ?
we can write fraction as the numerator and denominator so here the fraction of boys students is =  5/8
this fraction means and shows that there are 5 students are boys from 8 students.
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Alecsey [184]
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5 0
4 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
What is t?<br><br> 1/4t + 9 = 3
sertanlavr [38]
1/4t + 9 = 3
Subtract 9 from both sides
1/2t= -6
Divide 1/2 on both sides so that the only thing remaining on the left side is the variable t.
Final Answer: t= -12
6 0
3 years ago
Read 2 more answers
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