So lets try to prove it,
So let's consider the function f(x) = x^2.
Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).
Using the Intermediate Value Theorem,
it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.
Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.
Learn more about Intermediate Value Theorem on:
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Answer:
B happy to help!
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Step-by-step explanation:
Answer:
B) 11.8
Step-by-step explanation:
First write in 4 for x and 2 for y
Then multiply 1.3 times 6 (4+2) and subtract 2-6
Your expression should now be 7.8 +4
After adding these, it should be 11.8
Given:
Consider the given expression is
To find:
The radical form of given expression.
Solution:
We have,
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B2%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D%5Csqrt%5B5%5D%7By%5E3%7D)
![[\because x^{\frac{1}{n}}=\sqrt[n]{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%7D%5D)
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
Therefore, the required radical form is ![\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
.
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
<span> 5*(x-2)-4*(x+1)-(50)=0 </span>
