I hope this helps you
✔cos^2A+sin^2A=1
✔1-cos^2A=sin^2A
✔cos2A=cos^2A-sin^2A
✔sin2A=2.sinA.cosA
secA=1/cosA
tgA=sinA/cosA
sin^2A/1/cos^2A-sin^2A/cos^2A
sin^2A.cos^2A/cos2A
2.sin^2A.cos^2A/cos2A
sin2A.2.sin2A/cos2A
tg2A.2.sin2A
Answer:
The equivalent expression for the given expression
is
![4x^{3} y^{2}(\sqrt[3]{4xy} )](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%20%29)
Step-by-step explanation:
Given:
![\sqrt[3]{256x^{10}y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D)
Solution:
We will see first what is Cube rooting.
![\sqrt[3]{x^{3}} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20x)
Law of Indices

Now, applying above property we get
![\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%5Csqrt%5B3%5D%7B%284%5E%7B3%7D%5Ctimes%204%5Ctimes%20%28x%5E%7B3%7D%29%5E%7B3%7D%5Ctimes%20x%5Ctimes%20%28y%5E%7B2%7D%29%5E%7B3%7D%5Ctimes%20y%20%20%20%29%7D%20%5C%5C%5C%5C%5Ctextrm%7BCube%20Rooting%20we%20get%7D%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204%5Ctimes%20x%5E%7B3%7D%5Ctimes%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29%20%5C%5C%5C%5C%5Csqrt%5B3%5D%7B256x%5E%7B10%7Dy%5E%7B7%7D%20%7D%3D%204x%5E%7B3%7Dy%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%29)
∴ The equivalent expression for the given expression
is
![4x^{3} y^{2}(\sqrt[3]{4xy} )](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%28%5Csqrt%5B3%5D%7B4xy%7D%20%29)
Answer:
400 lb of salt
Step-by-step explanation:
Let us assume the water flows into the rank for x minutes.
There is an initial of 1000 gallons of water in the tank and water flows in through one pipe at 4 gal/min and through another pipe at 6 gal/min. In x minute, the amount of water in the tank = 1000 + 4x + 6x = 1000 + 10x
Water flows out at 5 gal/min, therefore in x minute the amount of water in the tank = 1000 + 10x - 5x = 1000 + 5x
The tank begins to overflow when it is full (has reached 1500 gallons). Therefore:
1500 = 1000 + 5x
5x = 1500 - 1000
5x = 500
x = 100 minutes.
1/2 lb salt per gallon flows into the tank at 4 gal/min and 1/3 lb of salt is flowing in at 6 gal/min, in 100 min the amount of salt that entered the tank = 4 gal/min × 100 min × 1/2 lb/gal + 6 gal/min × 100 min × 1/3 lb/gal= 400 lb
Therefore the amount of salt is in the tank when it is about to overflow = 400 lb of salt
Answer:
C : $58.36
Step-by-step explanation:
Given:
A store holiday sale has an item marked down by $10.
Discount on new price = 25%
Final price was $36.27.
Question asked:
what was the original ?
Solution:
Let original price = 
New price = 
Original price - marked down amount - discount amount = $36.27


Adding both side by 7.5

Dividing both side by 0.75

Therefore, the original price was $58.36