Answer:
A doctor sees more different patients each year than a teacher sees different students. The doctor can affect those people in ways that are much more personal and have much more of an immediate impact on their life. Teachers can help enrich the world intellectually, but doctors help keep us in good health.Both doctors and teachers are crucial professions. ... A doctor is one of the most important professions. After all, these specialists have knowledge and skills to diagnose, treat, and control the course of various diseases. Doctors save our lives.A teacher is better because he/she gives you the first advice on what to do and a doctor follows. Teachers are more knowledgeable than doctors. A teacher has knowledge about politicians, nurses and other professions yet doctors only treat patient. a teacher They encourage students to understand the importance of dedicating themselves to passion projects or endeavors. And success builds upon success. The more a student pushes themselves to accomplish their goals, the more they realize what they are capable of doing.
hope this helps :)
Option B is correct.
Step-by-step explanation:
We need to solve: ![\sqrt[3]{x^2}\sqrt[4]{x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E2%7D%5Csqrt%5B4%5D%7Bx%5E3%7D)
We know that: ![\sqrt[n]{x}\sqrt[b]{x} =\sqrt[n*b]{x.x}= \sqrt[n*b]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%5Csqrt%5Bb%5D%7Bx%7D%20%3D%5Csqrt%5Bn%2Ab%5D%7Bx.x%7D%3D%20%5Csqrt%5Bn%2Ab%5D%7Bx%5E2%7D)
Applying the above rule:
![\sqrt[3]{x^2}\sqrt[4]{x^3}\\=\sqrt[3*4]{x^2.x^3}\\=\sqrt[12]{x^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E2%7D%5Csqrt%5B4%5D%7Bx%5E3%7D%5C%5C%3D%5Csqrt%5B3%2A4%5D%7Bx%5E2.x%5E3%7D%5C%5C%3D%5Csqrt%5B12%5D%7Bx%5E5%7D)
So, Option B is correct.
Keywords: Solving with Exponents
Learn more about Solving with Exponents at:
#learnwithBrainly
Answer:
i tried but all i got was like 80 percent
Step-by-step explanation:
Answer:
w-2u-v
Step-by-step explanation:
Given are three vectors u, v and w.
In R^2 we treat first element as x coordinate and 2nd element as y coordinate.
Thus we mark (1,2) in the I quadrant, (-3,4) in II quadrant and (5,0) on positive x axis 5 units form the origin.
b) 
We have to find the values of a and b
]
Equate the corresponding terms
Divide II equation by 2 to get
Eliminate a
-5 = 5b: b=-1
a=-2
Hence
w = 2u-v
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>
