MATHEMATICS HIGH SCHOOL

Principal: $9,000, annual interest: 3%, interest periods: 12 , number of years: 13

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

P: the principal, amount invested

A: the new balance

t: the time

r: the rate, (in decimal form)

n: the number of times it is compounded.

Ex2:Suppose that $5000 is deposited in a saving account at the rate of 6% per year. Find the total amount on deposit at the end of 4 years if the interest is:

P =$5000, r = 6% , t = 4 years

a) simple : A = P(1+rt)

A = 5000(1+(0.06)(4)) = 5000(1.24) = $6200

b) compounded annually, n = 1:

A = 5000(1 + 0.06/1)(1)(4) = 5000(1.06)(4) = $6312.38

c) compounded semiannually, n =2:

A = 5000(1 + 0.06/2)(2)(4) = 5000(1.03)(8) = $6333.85

d) compounded quarterly, n = 4:

A = 5000(1 + 0.06/4)(4)(4) = 5000(1.015)(16) = $6344.93

e) compounded monthly, n =12:

A = 5000(1 + 0.06/12)(12)(4) = 5000(1.005)(48) = $6352.44

f) compounded daily, n =365:

A = 5000(1 + 0.06/365)(365)(4) = 5000(1.00016)(1460) = $6356.12

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If m < 0 and b > 0, the graph of y = mx + b does not pass through which quadrant?Quadrant I
Quadrant II
Quadrant III
Quadrant IV

Answers

Answer:

Quadrant III


Step-by-step explanation:

The attached picture shows graph of 4 such linear functions with the conditions given in the problem. ALL of them DO NOT pass through Quadrant III.

The graphs shown are of the functions:

y=-2x+1

y=-3x+3

y=-x+0.5

y=-5x+2


So, any linear function of the form  y=mx+b  with  m<0  and  b>0  does not pass through Quadrant III. Answer choice 3 is correct.
The graph will not past at quadrant III. I think thats the best answer, hope that helps.

A carpenter buys $54.38 worth of parts from supplier. If he pays for the parts with a $100 dollar bill, how much change will he receive?

Answers

$100 - $54.38 = $45.62

Answer:

45.62

Step-by-step explanation:

If a 100(degree) arc of a circle has a length of 8 inches, to the nearest inch, what is the radius of the circle?

Answers

 100(degree) arc of a circle has a length og 8 inches what is og

26.
What is the solution to. y - 9 > 4 + 2y?

HELP. answer if you can!​

Answers

Answer:

y<−13

y−9>4+2y

Simplify both sides of the inequality

y−9>2y+4

Subtract 2y from both sides

y−9−2y>2y+4−2y

−y−9>4

Add 9 to both sides.

−y−9+9>4+9

−y>13

The divide both sides by -1

Answer:

\boxed{y < -13}

Step-by-step explanation:

Hey there!

To solve for y we‘ll combine like terms and use the communicative property.

y - 9 > 4 + 2y

+9 to both sides

y > 13 + 2y

-2y to both sides

-y > 13

Divide -1 by both sides

y < -13

Hope this helps :)

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.

In a random sample of 75 individuals, 52 people said they prefer coffee to tea. 99.7% of the population mean is between 84/80/74 % and 64/58/54%. (Choose the option closest to your answer.)

Answers

Using a 1-sample z-interval for the population proportion with confidence level 0.997, the sample probability is 0.69 with margin of error of 0.16, so between 0.69+0.16=0.85 of the population and 0.69-0.16=0.53 of the population prefers coffee.  84% and 54% are the correct answers.

The correct answers are

84 and 54

:)
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