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steposvetlana [31]
3 years ago
5

If f(x)=3x-7,g(x)=x+2/5andg-1of(x)=f(x),find the value of x​

Mathematics
1 answer:
ad-work [718]3 years ago
5 0

Answer:

x = 5/2 = 2.5

Step-by-step explanation:

Given

f(x) = 3x-7

g(x) = x+2

g^(-1)(x) = f(x)

First find g^(-1)(x) = h(x)

x=h(x)+2

h(x) = x -2

h(x) = x - 2 = f(x) = 3x-7

solve for x

x-2  = 3x -7

-2 + 7 = 3x - x

2x = 5

x = 5/2

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What is the simplified form of 12x/900x²y4z6?
zalisa [80]

Answer:

360xy^2|xz^3|

Step-by-step explanation:

Given expression:

12x \sqrt{900x^2y^4z^6}

\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:

\implies 12x \sqrt{900}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}

Replace 900 with 30² :

\implies 12x \sqrt{30^2}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}

\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:

\implies 12x \cdot 30|x|\sqrt{y^4}\sqrt{z^6}

\implies 360x|x|\sqrt{y^4}\sqrt{z^6}

(We need to use the absolute value of √x² since the x term was originally to the power of 2, which means the value of x² is always positive since the exponent is even).

\textsf{Apply exponent rule} \quad \sqrt{a^m}=a^{\frac{m}{2}}:

\implies 360x|x|\cdot y^{\frac{4}{2}}\cdot z^{\frac{6}{2}}

Simplify:

\implies 360xy^2|xz^3|

(We need to use the absolute value of z³ since the z term was original to the power of 6, which means the value of z⁶ is always positive since the exponent is even).

7 0
2 years ago
How do I work this out I need full answer im lost​
taurus [48]

Answer:

7x = 35cm ; 2x = 10cm

Step-by-step explanation:

7x and 2x are the lengths

7x+2x=45

Then you take 7 + 2 = 9

so is gonna be 9x = 45

then you divide both side with 9

so is gonna be x = 5

Then you take 5 multiply 7 : 2

so 7 x 5 = 35cm

2 x 5 = 10cm

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3 years ago
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ASHA 777 [7]
C is the correct answer!
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4 years ago
How do I solve this triangle?
svetlana [45]

Answer:

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3 years ago
Estimate 76.216+66.36 by first rounding each number to the nearest whole number.
xeze [42]

Answer:

142

Step-by-step explanation:

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76.216=76\\66.36=66

  • Then ADD

76+66=142

6 0
3 years ago
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