MATHEMATICS HIGH SCHOOL

Cylinder has a height of 14 centimeters and its circle bases have a radius of 10 centimeters. find the surface area of the cylinder.2(100  ) + 140   square centimeters
1,400  + 140   square centimeters
2(100  ) + 280   square centimeters
2(1,400  ) + 2(140  )  square centimeters

Answers

Answer 1
Answer: 200 plus 140 is 300 
1400 plus 140 is 1540
200 plus 280 is 480
and 2800 plus 280 3080

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Which of the following statements is always true?a.
Workers being paid on commission make less money than if they are salaried.
b.
Workers being paid on commission get paid based solely on their performance.
c.
Workers being paid on commission are stressed over the amount of earnings they will have.
d.
Workers being paid on commission increase the accounting costs of the employer.

Answers

Workers being paid on commission get paid based solely on their performance.
They get paid based on their performance and not their hours.

Answer:

its b

Step-by-step explanation:

The resquest of the y'+6y=e^4t, y(0)=2

Answers

This is a linear differential equation of first order. Solve this by integrating the coefficient of the y term and then raising e to the integrated coefficient to find the integrating factor, i.e. the integrating factor for this problem is e^(6x). 
Multiplying both sides of the equation by the integrating factor: 

(y')e^(6x) + 6ye^(6x) = e^(12x) 

The left side is the derivative of ye^(6x), hence 

d/dx[ye^(6x)] = e^(12x) 

Integrating 

ye^(6x) = (1/12)e^(12x) + c where c is a constant 

y = (1/12)e^(6x) + ce^(-6x) 

Use the initial condition y(0)=-8 to find c: 

-8 = (1/12) + c 
c=-97/12 

Hence 

y = (1/12)e^(6x) - (97/12)e^(-6x)

A rectangular plot of ground is 5 meters longer than it is wide. Its area is 20,000 square meters.which equation will help you find dimensions?
Let w=width
a- w+2(w+5)=20,000
b- w(w+5)=20,000
c- (w(w+5))/2 = 20,000
d- w^2=20,000+5

Answers

Answer:

b- w(w+5)=20,000

Step-by-step explanation:

please mark this answer as brainliest

Show how the greateat common factor of the number 10 and 15 can be used to reduce the fraction 10/15

Answers

The greatest common factor of 10 and 15 is 5. To find that answer, start by listing the factors of both numbers.

10: 1, 2, 5, 10
15: 1, 3, 5, 15

The GCF is 5. Using this, we can divide both numbers by 5 to get 2/3, which is the simplified version of 10/15.

Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x)How about cos(x) and tan(x)?

Answers

1.

f'(\sin x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\sin(x+h) - \sin(x))/(h) = \n \n = \lim_(h \to 0) (2 \sin( (x+h - x)/(2)) \cdot \cos( (x+h+x)/(2)) )/(h) = \lim_(h \to 0) (2 \sin( (h)/(2)) \cos( (2x+h)/(2) ) )/(h) = \n \n = \lim_(h \to 0) [ (\sin( (h)/(2)) )/( (h)/(2) ) \cdot \cos ((2x+h)/(2)) ] = \lim_(h \to 0) [1 \cdot \cos( (2x+h)/(2) ) ] =

= \cos( (2x)/(2)) = \boxed{\cos x}

2.

f'(\cos x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cos(x+h) - \cos(x))/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (x+h+x)/(2)) \cdot \sin ( (x+h-x)/(2)) )/(h) = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) \cdot \sin ( (h)/(2)) )/(h) = \n \n = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) )/(2) \cdot (sin( (h)/(2)) )/( (h)/(2) ) = \lim_(h \to 0) -\sin( (2x+h)/(2)) \cdot 1 =

= -\sin( (2x)/(2)) = \boxed{\sin x }

3.

f'(\tan) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\tan(x+h) - \tan(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x+h-x))/(\cos(x+h) \cdot \cos(x)) )/(h) = \lim_(h \to 0) ( (\sin(h))/( (\cos(x+h-x) + \cos(x+h+x))/(2) ) )/(h) =

= \lim_(h \to 0) ( (\sin(h))/(\cos(h) + \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) (\sin(h))/( (1)/(2)h \cdot [\cos(h) + \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (\sin(h))/(h) \cdot (1)/( (1)/(2) \cdot (\cos(h) + cos(2x+h) ) = 1 \cdot (1)/( (1)/(2) \cdot (1+ cos(2x) ) = (2)/(1 + 2 \cos^(2) - 1 ) = \n \n = (2)/(2 \cos^(2) x) = \boxed{ (1)/(\cos^(2)x) }

4.

f'(\cot) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cot(x+h) - \cot(x))/(h) = \n \n = \lim_(h \to 0) ( (\sin(x - x - h))/(\sin (x+h) \cdot \sin (h)) )/(h) = \lim_(h \to 0) ( (\sin(-h) )/( (\cos(x+h-x) - \cos(x+h+x))/(2) ) )/(h) =

= \lim_(h \to 0) ( (-\sin(h))/(\cos(h) - \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) ( - \sin(h))/( (1)/(2)h \cdot [\cos(h) - \cos(2x+h)] ) = \n \n = \lim_(h \to 0) (- \sin (h))/(h) \cdot (1)/( (1)/(2) \cdot [\cos(h) - \cos(2x+h)] ) = -1 \cdot (2)/(1 - cos(2x)) = \n \n = - (2)/(1 -1 + 2 \sin^(2)x) = - (2)/(2 \sin^(2) x) = \boxed{- (1)/(\sin^(2) x) }
I posted an image instead.

Highest common multiple of 8 and 14

Answers

There is no specified highest common multiple of two numbers, because a number has an infinite number of multiples extending out to infinity. So, if anything, the highest common multiple of 8 and 14 is infinity.
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