Volume of piece candy is 113.04 cubic centimeter
<em><u>Solution:</u></em>
A student bought a piece of hard candy that is in the shape of a sphere
<em><u>The candy has a radius of 3 centimeters</u></em>
Radius = 3 cm
Use 3.14 for π
<em><u>Volume of sphere is given as:</u></em>
![V = \frac{4}{3} \times \pi \times r^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%20%5Cpi%20%5Ctimes%20r%5E3)
Where, "r" is the radius of sphere
<em><u>Therefore, volume of candy is given as:</u></em>
![V = \frac{4}{3} \times 3.14 \times 3^3\\\\V = 4 \times 3.14 \times 3^2\\\\V = 12.56 \times 9\\\\V = 113.04](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%203.14%20%5Ctimes%203%5E3%5C%5C%5C%5CV%20%3D%204%20%5Ctimes%203.14%20%5Ctimes%203%5E2%5C%5C%5C%5CV%20%3D%2012.56%20%5Ctimes%209%5C%5C%5C%5CV%20%3D%20113.04)
Thus volume of candy is 113.04 cubic centimeter
That means its another way scientist use this method to easly use large numbers or very small numbers for example 0.0000000056, they make it like this 5.6*10 to the -9 power
Hope it helps to solve it
Answer:
Step-by-step explanation:
The A
Answer:
Step-by-step explanation:
Hello,
as cosx+sinx=k we can write that
![(cosx+sinx)^2=cos^2x+sin^2x+2\ cosx \ sinx = k^2](https://tex.z-dn.net/?f=%28cosx%2Bsinx%29%5E2%3Dcos%5E2x%2Bsin%5E2x%2B2%5C%20cosx%20%5C%20sinx%20%3D%20k%5E2)
and we know that ![cos^2x+sin^2x=1](https://tex.z-dn.net/?f=cos%5E2x%2Bsin%5E2x%3D1)
so
![2 \ cosx \ sinx \ = k^2 -1 \\ sinx \ cosx = \dfrac{k^2-1}{2}](https://tex.z-dn.net/?f=2%20%5C%20cosx%20%5C%20sinx%20%5C%20%3D%20k%5E2%20-1%20%5C%5C%3C%3D%3E%20sinx%20%5C%20cosx%20%3D%20%5Cdfrac%7Bk%5E2-1%7D%7B2%7D)
which is the answer to the question 2
Now, let s estimate
![1=1^2=(cos^2x+sin^2x)^2=cos^4x+sin^4x+2cos^2xsin^2x](https://tex.z-dn.net/?f=1%3D1%5E2%3D%28cos%5E2x%2Bsin%5E2x%29%5E2%3Dcos%5E4x%2Bsin%5E4x%2B2cos%5E2xsin%5E2x)
so
![cos^4x+sin^4x=1-2cos^2xsin^2x](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D1-2cos%5E2xsin%5E2x)
We use the previous result to write
![cos^4x+sin^4x=1-2cos^2xsin^2x = 1-2(\dfrac{k^2-1}{2})^2 = \dfrac{2-(k^2-1)^2}{2}](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D1-2cos%5E2xsin%5E2x%20%3D%201-2%28%5Cdfrac%7Bk%5E2-1%7D%7B2%7D%29%5E2%20%3D%20%5Cdfrac%7B2-%28k%5E2-1%29%5E2%7D%7B2%7D)
and we know that
![(k^2-1)^2=k^4-2k^2+1](https://tex.z-dn.net/?f=%28k%5E2-1%29%5E2%3Dk%5E4-2k%5E2%2B1)
so
![cos^4x+sin^4x=\dfrac{2-k^4+2k^2-1}{2}=\dfrac{-k^4+2k^2+1}{2}](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D%5Cdfrac%7B2-k%5E4%2B2k%5E2-1%7D%7B2%7D%3D%5Cdfrac%7B-k%5E4%2B2k%5E2%2B1%7D%7B2%7D)
this is the answer to the first question
finally, let s estimate
so ![(sinx-cosx)=\sqrt{2-k^2}](https://tex.z-dn.net/?f=%28sinx-cosx%29%3D%5Csqrt%7B2-k%5E2%7D)
and this is the answer to the last question
do not hesitate if you need further explanation
hope this helps