You can do this by comparing your equation to the general form of equation of a circle
( x - a)^2 + (y - b)^2 = r^2 where (a,b) is the centre and r = radius:-
(x + 9)^2 (y + 6)^2 = 9
so (a,b) = (-9,-6) and r = 3
so the answer is B
An exponential function displays either a growth or decay behavior going from a low steep to a high steep.
The equation that could be solved to find x, the measure of AC is 58 = 1/2(238 -x)
<h3>Circle theorem</h3>
The given diagram shows two intersecting lines tangential to a circle at points A and C.
Using the theorem that states, the measure of the angle at the vertex is equal to the half of the difference of the measure of the intercepted arcs.
Mathematically;
<B = 1/2(arcADC - arcAC)
58 = 1/2(238 -x)
Hence the equation that could be solved to find x, the measure of AC is 58 = 1/2(238 -x)
Learn more on circle theorem here: brainly.com/question/26594685
#SPJ1
Answer:
![1. \quad\dfrac{1}{k^{\frac{2}{3}}}\\\\2. \quad\sqrt[7]{x^5}\\\\3. \quad\dfrac{1}{\sqrt[5]{y^2}}](https://tex.z-dn.net/?f=1.%20%5Cquad%5Cdfrac%7B1%7D%7Bk%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%5C%5C%5C%5C2.%20%5Cquad%5Csqrt%5B7%5D%7Bx%5E5%7D%5C%5C%5C%5C3.%20%5Cquad%5Cdfrac%7B1%7D%7B%5Csqrt%5B5%5D%7By%5E2%7D%7D)
Step-by-step explanation:
The applicable rule is ...
![x^{\frac{m}{n}}=\sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%5Em%7D)
It works both ways, going from radicals to frational exponents and vice versa.
The particular power or root involved can be in either the numerator or the denominator. The transformation applies to the portion of the expression that is the power or root.
Answer:
![\sqrt[3]{0.95} \approx 0.9833](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.95%7D%20%5Capprox%200.9833)
![\sqrt[3]{1.1} \approx 1.0333](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1.1%7D%20%5Capprox%201.0333)
Step-by-step explanation:
Given the function: ![g(x)=\sqrt[3]{1+x}](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%5B3%5D%7B1%2Bx%7D)
We are to determine the linear approximation of the function g(x) at a = 0.
Linear Approximating Polynomial,
a=0
![g(0)=\sqrt[3]{1+0}=1](https://tex.z-dn.net/?f=g%280%29%3D%5Csqrt%5B3%5D%7B1%2B0%7D%3D1)
Therefore:
(b)![\sqrt[3]{0.95}= \sqrt[3]{1-0.05}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.95%7D%3D%20%5Csqrt%5B3%5D%7B1-0.05%7D)
When x = - 0.05
(c)
(b)![\sqrt[3]{1.1}= \sqrt[3]{1+0.1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1.1%7D%3D%20%5Csqrt%5B3%5D%7B1%2B0.1%7D)
When x = 0.1
