I think 33.3% that he Chooses that
Answer:
=33
Step-by-step explanation:
Answer: l = 4 cm, w = 7 cm
Step-by-step explanation:
i think that this is it
Answer:
See below
Step-by-step explanation:
Here we need to prove that ,
![\sf\longrightarrow sin^2\theta + cos^2\theta = 1](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20sin%5E2%5Ctheta%20%2B%20cos%5E2%5Ctheta%20%3D%201%20)
Imagine a right angled triangle with one of its acute angle as
.
- The side opposite to this angle will be perpendicular .
- Also we know that ,
![\sf\longrightarrow sin\theta =\dfrac{p}{h} \\](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20sin%5Ctheta%20%3D%5Cdfrac%7Bp%7D%7Bh%7D%20%5C%5C)
![\sf\longrightarrow cos\theta =\dfrac{b}{h}](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20cos%5Ctheta%20%3D%5Cdfrac%7Bb%7D%7Bh%7D%20)
And by Pythagoras theorem ,
![\sf\longrightarrow h^2 = p^2+b^2 \dots (i)](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20h%5E2%20%3D%20p%5E2%2Bb%5E2%20%5Cdots%20%28i%29%20)
Where the symbols have their usual meaning.
Now , taking LHS ,
![\sf\longrightarrow sin^2\theta +cos^2\theta](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20sin%5E2%5Ctheta%20%2Bcos%5E2%5Ctheta%20)
- Substituting the respective values,
![\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20%5Cbigg%28%5Cdfrac%7Bp%7D%7Bh%7D%5Cbigg%29%5E2%2B%5Cbigg%28%5Cdfrac%7Bb%7D%7Bh%7D%5Cbigg%29%5E2%5C%5C)
![\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20%5Cdfrac%7Bp%5E2%7D%7Bh%5E2%7D%2B%5Cdfrac%7Bb%5E2%7D%7Bh%5E2%7D%5C%5C%20)
![\sf\longrightarrow \dfrac{p^2+b^2}{h^2}](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20%5Cdfrac%7Bp%5E2%2Bb%5E2%7D%7Bh%5E2%7D%20)
![\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%5Ccancel%7B%20%5Cdfrac%7Bh%5E2%7D%7Bh%5E2%7D%7D%5C%5C%20)
![\sf\longrightarrow \bf 1 = RHS](https://tex.z-dn.net/?f=%5Csf%5Clongrightarrow%20%5Cbf%201%20%3D%20RHS%20)
Since LHS = RHS ,
Hence Proved !
I hope this helps.
Answer:
Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.
Step-by-step explanation:
The equation f(x) given is:
![f(x) = x^3+6x-5](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E3%2B6x-5)
For x = 0. the value of the expression is:
![f(0) = 0^3+0-5\\f(0) = -5](https://tex.z-dn.net/?f=f%280%29%20%3D%200%5E3%2B0-5%5C%5Cf%280%29%20%3D%20-5)
For x = 1, the value of the expression is:
![f(1) = 1^3+6-5\\f(1)=2](https://tex.z-dn.net/?f=f%281%29%20%3D%201%5E3%2B6-5%5C%5Cf%281%29%3D2)
Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.
In other words, there is at least one solution for the equation between x=0 and x=1.