That will depend, you have to ask yourself first how many kilobytes one picture is. Let's just say that the size of one picture is 100 kb (which is the average size of a picture) Then first you multiply 100 kb with the number of pictures which is 43. Now you have a total used up memory of 4300 kb. After that, you minus the used up memory which is 4300 kb, to the total available space which is 32,834.5 and you will get an available space of 28534.5 kb. After that, you divide the remaining available space with the size of each picture. So this will be 28534.5 divided by 100. You will get 285. You can still take 285 pictures.
I'm just guessing but maybe expressions
Answer:
Step-by-step explanation:
cos(c)= x/ac
ac = cos32/9.4
ac =11.08
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.