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1 year ago

Arithmetic and Geometric Sequences (Context)

Mathematics

1 answer:

V125BC [204]1 year ago

7 0

The formula for compound interest

A = P( 1 + r/n) ^ (nt)

A is the amount in the account at the end

P is the principal balance or the amount initially invested

r is the annual interest rate in decimal form

n is the number of times it is coupounded per year

t is the number of years

A = 1800 ( 1+ .0375/1) ^ (1*6)

A = 1800 ( 1.0375)^6

A = 2244.92138

Rounding to the nearest cent

A = 2244.92

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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

6 0

3 years ago

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Enter the variable that is easiest to solve for in this system of equations. 6y=10x+5 , 6y=5x+7

djverab [1.8K]

Answer:

Step-by-step explanation:

nhmnj

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3 years ago

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Pythagoras was born about 582 bc. Isaac Newton was born in 1643 ad. How many years apart were they born?

malfutka [58]

B.C. counts down to 0, and that's when A.D. starts counting up. That means you can add 582 and 1643, to get their difference. Pythagoras and Newton were born 2,225 years apart.

8 0

3 years ago

Round to the nearest tenth :14.94 need answer ASAP, please explain your answer!

Diano4ka-milaya [45]

14.9

Step-by-step explanation:

Here’s a trick 5 or more let it sore 4 or less let it rest.

4 0

2 years ago

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Use the Shell Method to find the volume of the solid obtained by rotating region under the graph of f(x)=x^2+2 for 0≤x≤5 about x

Irina18 [472]

For each x in the interval 0 ≤ x ≤ 5, the shell at that point has

• radius = 5 - x, which is the distance from x to x = 5

• height = x ² + 2

• thickness = dx

and hence contributes a volume of 2π (5 - x) (x ² + 2) dx.

Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:

$\displaystyle 2\pi \int_0^5 (5-x)(x^2+2)\,\mathrm dx=2\pi\int_0^5 (10-2x+5x^2-x^3)\,\mathrm dx=\boxed{\frac{925\pi}6}$

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2 years ago