Answer:
-3p+21≥21
Step-by-step explanation:
<h3>Given Equation:</h3>
4x + 9 = 56
<h3>To Find:</h3>
The value of x.
<h3>Solution:</h3>
4x + 9 = 56
The value of 9 changes when we take 9 to the Right Hand Side, we get
4x = 56 - 9
Now, The value of 4 changes when we take 4 to the Right Hand Side, we get
x = 47/4
<h2>Answer:</h2>
The value of x is 47/4
Step-by-step explanation:
- The square root of a product of two positive real numbers is the product of their square roots.
Let 4 and 5 be the two positive real number. The square root of a product of 4 and 5 is the product of their square roots.



![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D)

so

![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da)

Therefore,

NOW SOLVING BY APPLYING RADICAL RULE such that
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D)

as

so
Hence, The square root of a product of two positive real numbers is the product of their square roots.
Answer:
![\[3x+\frac{2}{3}\]](https://tex.z-dn.net/?f=%5C%5B3x%2B%5Cfrac%7B2%7D%7B3%7D%5C%5D)
Step-by-step explanation:
![\[f(x)=x-\frac{1}{3}\]](https://tex.z-dn.net/?f=%5C%5Bf%28x%29%3Dx-%5Cfrac%7B1%7D%7B3%7D%5C%5D)
![\[g(x)=3x+1\]](https://tex.z-dn.net/?f=%5C%5Bg%28x%29%3D3x%2B1%5C%5D)
Hence, ![\[(f o g)(x)=f(3x+1)\]](https://tex.z-dn.net/?f=%5C%5B%28f%20o%20g%29%28x%29%3Df%283x%2B1%29%5C%5D)
But, ![\[f(3x+1)=(3x+1)-\frac{1}{3}\]](https://tex.z-dn.net/?f=%5C%5Bf%283x%2B1%29%3D%283x%2B1%29-%5Cfrac%7B1%7D%7B3%7D%5C%5D)
Simplifying,
![\[f(3x+1)=3x+(1-\frac{1}{3})\]](https://tex.z-dn.net/?f=%5C%5Bf%283x%2B1%29%3D3x%2B%281-%5Cfrac%7B1%7D%7B3%7D%29%5C%5D)
= ![\[f(3x+1)=3x+(\frac{3-1}{3})\]](https://tex.z-dn.net/?f=%5C%5Bf%283x%2B1%29%3D3x%2B%28%5Cfrac%7B3-1%7D%7B3%7D%29%5C%5D)
= ![\[f(3x+1)=3x+(\frac{2}{3})\]](https://tex.z-dn.net/?f=%5C%5Bf%283x%2B1%29%3D3x%2B%28%5Cfrac%7B2%7D%7B3%7D%29%5C%5D)
Hence, ![\[(f o g)(x)=3x+(\frac{2}{3})\]](https://tex.z-dn.net/?f=%5C%5B%28f%20o%20g%29%28x%29%3D3x%2B%28%5Cfrac%7B2%7D%7B3%7D%29%5C%5D)
Step-by-step explanation:
take into 2 sections
-5<a-4 and a-4<2
-1<a and a<6
therefore it should be the 2nd one