Which number is irrational, an integer and a real number?

1 answer:

8 0

Hello!

If a number is irrational it is not an integer but could be a real number. But since a number cannot be both irrational and an integer, our answer is D) there is no such number.

I hope this helps!

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Veronika [31]

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.. f(x) = x²
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.. g(x) = 3x²

6 0

3 years ago

Find the value 'X' in this given picture

taurus [48]

Hi.
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4 years ago

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Solve the following exponential equations.<br> 2(4^x )+ 4^(x+1 )= 342

Cloud [144]

Answer:

x ≈ 2.91644500708

Step-by-step explanation:

The equation can be simplified to ...

2(4^x) +4(4^x) = 342

6(4^x) = 342

4^x = 57

Taking logarithms, we get ...

x = log₄(57) = log(57)/log(4)

x ≈ 2.91644500708

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%7D%20%287%5Csqrt%5B3%5D%7B8y%5E2%7D-%5Csqrt%5B3%5D%7By%5E5%7D%20-4y%5Csqrt%

noname [10]

Answer:

$\huge\boxed{-y^2+2y}$

Step-by-step explanation:

$\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y$