Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
Answer:
2
Step-by-step explanation:
Step-by-step explanation:
4x + 2xy=
4×4 + 2×4×-3
= 16 + (-24)
= - 8
<h2>
Hello!</h2>
The answers are:
The possible values for x in the equation, are:
First option, ![5\sqrt[3]{3}](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7B3%7D)
Second option, ![\sqrt[3]{375}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B375%7D)
<h2>
Why?</h2>
To solve the problem, we need to remember the following properties of the exponents and roots:
![a\sqrt[n]{b}=\sqrt[n]{a^{n}*b} \\\\\sqrt[n]{a^{m} }=a^{\frac{m}{n}}\\\\(a^{b})^{c}=a^{b*c}](https://tex.z-dn.net/?f=a%5Csqrt%5Bn%5D%7Bb%7D%3D%5Csqrt%5Bn%5D%7Ba%5E%7Bn%7D%2Ab%7D%20%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%20%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%5C%5C%5C%5C%28a%5E%7Bb%7D%29%5E%7Bc%7D%3Da%5E%7Bb%2Ac%7D)
Then, we are given the expression:
So, finding "x", we have:
![x^{3}=375\\\\(x^{3})^{\frac{1}{3} } =(375)^{\frac{1}{3}}\\\\x=\sqrt[3]{375}=\sqrt[3]{125*3}=\sqrt[3]{125}*\sqrt[3]{3}=5\sqrt[3]{3}](https://tex.z-dn.net/?f=x%5E%7B3%7D%3D375%5C%5C%5C%5C%28x%5E%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%28375%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B375%7D%3D%5Csqrt%5B3%5D%7B125%2A3%7D%3D%5Csqrt%5B3%5D%7B125%7D%2A%5Csqrt%5B3%5D%7B3%7D%3D5%5Csqrt%5B3%5D%7B3%7D)
Hence, the possible values for x in the equation, are:
First option,
Second option,
Have a nice day!
