Answer:I love thee fact that I am a straight a Student but I don't wanna help you dumb people
Step-by-step explanation:
Answer:17
Step-by-step explanation:
Answer: 22
explanation: the easiest way is to separate one of the diagonals into a triangle and use the pythagorean theorem.
a^2 + b^2 = c^2
4^2 + 3^3 = c^2
16 + 9 = c^2
25 = c^2
5 = c
you now know that both of the diagonals have a length of 5.
by counting the units on the two straight, you know that their length is 6.
6 + 6 + 5 + 5 = 22
Answer:
T+V/I
Step-by-step explanation:
multiply both sides by I to get rid of fractions ti=V
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.