A savings account accrues interest at a rate of 3.0% yearly. If someone opens an account with $2,500, how much money would the account have after 5 years?

Answers

Answer 1

Answer: $3\%=(3)/(100)=0,03\n\n First\ year:2500+2500*0,03=2500+75=2575\n\nSecond\ year:\ 2575+2575*0,03=2575+77,25=2652,25\n\nThird\ year:\ \ 2652,25+2652,25*0,03=2652,25+79,5675=\n\n2731,8175\n\n Fourth\ year: 2731,8175+2731,8175*0,03=2731,8175+81,954525\n\n=2813,772025\n\nFifth\ year:\ 2813,772025+2813,772025*0,03=\n\n2813,772025+84,41316075=2898,18518575\$$

Related Questions

What is the square root of 300 rounded to the nearest tenth?
270 is 75% of what number?
How many pence in £3.05
Andrea is shopping for items for her pet sitting business because the store is having a sale where everything in the store is 5% off. Andrea buys a cat carrier for $39.99 and a cat tower for $19.99. She has a coupon for an additional 15% off on the cat tower. Before tax, Andrea estimates that her total, after discounts, will be about $54. Determine if Andrea has estimated correctly.
Explain how you would write 36/10 as s decimal

Carlota has 3/4 ton of mulch she is going to divide evenly among 5 flower beds. How much mulch will each flower bed contain? Please show work

Answers

The first thing to do is to turn 5 into a fraction.
5 = 5/1

Now it is just a simple fractional division question:
3/4 / 5/1

With fractional division, all you do is switch around the fraction that is on the right and turn the divide sign into a multiplication sign:
3/4 X 1/5

Now you simply multiply the numerators together, and the denominators together:
3 X 1 = 3
4 X 5 = 20

3/20.

Each flower bed will contain 3/20 of a ton of mulch.

Express 53 = x as a logarithmic equation.

Answers

Answer:

$\log_5 x = 3$

Step-by-step explanation:

Using logarithmic rule:

$\log_b b^a = a$

Given the equation:

$5^3 = x$

Taking log with base 5 both sides we have;

$\log_5 5^3 = \log_5 x$

Apply the logarithmic rule:

$3 = \log_5 x$

$\log_5 x = 3$

Therefore, $5^3 = x$ as a logarithmic equation is, $\log_5 x = 3$

log5x=3 I think is the answer

carla has two lengths of ribbon. one ribbon is 2 feet long. the other ribbon is 30 inches long.which length of ribbon is longer?explain.

Answers

1 ft = 12 in
2 ft = 24 in
So the 2nd ribbon is larger

1 foot = 12 inches
2 feet = 24 inches

24 < 30
so the ribbon that is 30 inches is longer

About how far can a subway car travel in 1 hour if its average is 0.96 miles per minute?

Answers

Answer:

57.6 miles per hour

Step-by-step explanation:

0.96 miles per minute

60 minutes per hour

0.96X60=57.6 miles per hour

57.6 is the answer. have a great dayyyyy :)

12 1/4 - 5 2/3 subtract fraction with remaining

Answers

The value of the fraction is given by equation A = 6 7/12

What is a Fraction?

An element of a whole is a fraction. The number is represented mathematically as a quotient, where the numerator and denominator are split. Both are integers in a simple fraction. A fraction appears in the numerator or denominator of a complex fraction. The numerator of a proper fraction is less than the denominator.

Given data ,

Let the simplified fraction be represented as A

Now , let the first fraction be p

The value of p = 12 1/4

Let the second fraction be q

The value of q = 5 2/3

On simplifying , we get

A = p - q

A = 12 1/4 - 5 2/3

A = ( 49/4 ) - ( 17/3 )

Taking the LCM of the denominators , we get

A = [ ( 49 x 3 ) - ( 17 x 4 ) ] / 12

A = ( 147 - 68 ) / 12

A = ( 79/12 )

On further simplification of the fraction , we get

A = 6 7/12

Therefore , the value of A is 6 7/12

Hence , the simplified fraction is A = 6 7/12

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12 1/4 = 49/4
5 2/3 = 17/3

49/4 = 147/12
17/3 = 68/12

14/12 - 68/12 = 79/12
6 5/12

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.x2 + y2 − 2x + 2y − 1 = 0
x2 + y2 − 4x + 4y − 10 = 0
x2 + y2 − 8x − 6y − 20 = 0
4x2 + 4y2 + 16x + 24y − 40 = 0
5x2 + 5y2 − 20x + 30y + 40 = 0
2x2 + 2y2 − 28x − 32y − 8 = 0
x2 + y2 + 12x − 2y − 9 = 0

Answers

The correct answer is:

x²+y²-2x+2y-1 = 0;
x²+y²-4x+4y-10 = 0;
5x²+5y²-20x+30y+40 = 0;
x²+y²-8x-6y-20 = 0;
x²+y²+12x-2y-9 = 0;
4x²+4y²+16x+24y-40 = 0; and
2x²+2y²-28x-32y-8 = 0

Explanation:

For each of these, we want to write the equation in the form
(x+h)²+(y+k)² = r².

To do this, we evaluate the terms 2hx and 2ky in each equation. We will take half of this; this will tell us what h and k are for each equation.

For the first equation:
2hx = -2x and 2ky = 2y.

Half of -2x = -1x and half of 2y = 1y; this means h = -1 and k = 1:
(x-1)² + (y+1)² + ___ - 1 = 0

When we multiply (x-1)², we get
x²-2x+1.
When we multiply (y+1)², we get
y²+2y+1.

This gives us 1+1 = 2 for the constant. We know we must add something to 2 to get -1; 2 + ___ = -1; the missing term is -3. Add that to each side (to have r² on the right side of the equals) and we have
(x-1)² + (y+1)² = 3
This means that r² = 3, and r = √3 = 1.732.

For the second equation, 2hx = -4x and 2ky = 4y; this means h = -4/2 = -2 and k = 4/2 = 2. This gives us
(x-2)² + (y+2)² -10 + ___ = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+2)² gives us
y²+4x+4.
This gives us 4+4= 8 for our constant so far.

We know 8 + ___ = -10; this means the missing term is -18. Add this to each side of the equation to have
(x-2)²+(y+2)² = 18; r² = 18; r = √18 = 3√2 = 4.243.

For the third equation, 2hx = -8x and 2ky = -6y. This means h = -8/2 = -4 and k = -6/2 = -3. This gives us:
(x-4)²+(y-3)²-20 = 0

Multiplying (x-4)² gives us
x²-8x+16.
Multiplying (y-3)² gives us
y²-6y+9.

This gives us 16+9 = 25 for the constant. We know that 25+___ = -20; the missing term is -45. Add this to each side for r², and we have that
r²=45; r = √45 = 3√5 = 6.708.

For the next equation, we factor 4 out of the entire equation:
4(x²+y²+4x+6y-10)=0.
This means 2hx = 4x and 2ky = 6y; this gives us h = 4/2 = 2 and k = 6/2 = 3. This gives us
4((x+2)²+(y+3)² - 10) = 0.

Multiplying (x+2)² gives us
x²+4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know 13+__ = -10; this missing value is -23. Since we had factored out a 4, that means we have 4(-23) = -92. Adding this to each side for r², we have
r²=92; r = √92 = 2√23 = 9.59.

For the next equation, we factor out a 5 first:
5(x²+y²-4x+6y+8) = 0. This means that 2hx = -4x and 2ky = 6y; this gives us h = -4/2 = -2 and k = 6/2 = 3:

5((x-2)²+(y+3)²+8) = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know that 13+__ = 8; the missing value is -5. Since we factored a 5 out, we have 5(-5) = -25. Adding this to each side for r² gives us
r²=25; r = √25 = 5.

For the next equation, we first factor a 2 out:
2(x²+y²-14x-16y-4) = 0. This means 2hx = -14x and 2ky = -16y; this gives us h = -14/2 = -7 and k = -16/2 = -8:

2((x-7)²+(y-8)²-4) = 0.

Multiplying (x-7)² gives us
x²-14x+49.
Multiplying (y-8)² gives us
y²-16x+64.

This gives us a constant of 49+64=113. We know that 113+__ = -4; the missing value is -117. Since we first factored out a 2, this gives us 2(-117) = -234. Adding this to each side for r² gives us
r²=234; r = √234 = 3√26 = 15.297.

For the last equation, 2hx = 12x and 2ky = -2; this means h = 12/2 = 6 and k = -2/2 = -1:
(x+6)²+(y-1)²-9 = 0

Multiplying (x+6)² gives us
x²+12x+36.
Multiplying (y-1)² gives us
y²-2y+1.

This gives us a constant of 36+1 = 37. We know that 37+__ = -9; the missing value is -46. Adding this to each side for r² gives us
r² = 46; r=√46 = 6.78.

Find the radius of each equation:

1.
$x^2 + y^2-2x+2y-1 = 0, \n x^2-2x+1-1 + y^2+2y+1-1-1 = 0, \n (x-1)^2+(y+1)^2=3$ , then $r_1= √(3)$ .

2.
$x^2 + y^2-4x + 4y- 10 = 0, \n x^2 -4x+4-4+ y^2 + 4y+4-4- 10 = 0, \n (x-2)^2+(y+2)^2=18$ , then $r_2= √(18)=3 √(2)$ .

3.
$x^2 + y^2-8x- 6y- 20 = 0, \n x^2-8x+16-16+ y^2- 6y+9-9- 20 = 0, \n (x-4)^2+(y-3)^2=45$ , then $r_3= √(45) =3 √(5)$ .

4.
$4x^2 + 4y^2+16x+24y- 40 = 0, \n 4x^2+16x+16-16+ 4y^2+24y+36-36- 40 = 0, \n 4(x+2)^2+4(y+3)^2=92,\n (x+2)^2+(y+3)^2=23$ , then $r_4= √(23)$ .

5.
$5x^2 + 5y^2-20x+30y+ 40 = 0, \n 5x^2-20x+20-20+ 5y^2+30y+45-45- 40 = 0, \n 5(x-2)^2+5(y+3)^2=105,\n (x-2)^2+(y+3)^2=21$ , then $r_5= √(21)$ .

6.
$2x^2 + 2y^2-28x-32y- 8= 0, \n 2x^2-28x+98-98+ 2y^2-32y+128-128- 8= 0, \n 2(x-7)^2+2(y-8)^2=234,\n (x+2)^2+(y+3)^2=117$ , then $r_6= √(117)=3√(13)$ .

7.
$x^2 + y^2+12x-2y-9 = 0, \n x^2+12x+36-36+ y^2-2y+1-1- 9 = 0, \n (x+6)^2+(y-1)^2=46$ , then $r_7= √(46)$ .

Hence
$r_1= √(3)$ , $r_2=3 √(2)$ , $r_3=3 √(5)$ , $r_4= √(23)$ , $r_5= √(21)$ , $r_6= 3√(13)$ , $r_7= √(46)$ and $r_1\ \textless \ r_2\ \textless \ r_5\ \textless \ r_4\ \textless \ r_3\ \textless \ r_7\ \textless \ r_6$ .

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