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2 years ago

Suppose $500 is invested at 6% annual interest compounded twice a year. When will the investment be worth $1000.

Mathematics

1 answer:

Nataly_w [17]2 years ago

7 0

Answer:

Step-by-step explanation:

$V=P(1+r/c)^{(ct)}\\ \\ V=P(1+.06/2)^{(2t)}\\ \\ 1000=500(1.03)^{(2t)}\\ \\ (1.03)^{2t}=2\\ \\ 2t=ln2/(ln1.03)\\ \\ t=ln2/(2ln(1.03))\\ \\ t\approx 11.7 years$

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Solve the equation by factoring. 2x^2 + 15x + 25 = 0

Pepsi [2]

Answer:(2x+5)(x+5)=0

Step-by-step explanation:

2x^2+15x+25=0

Factorise we have

2x^2+10x+5x+25

(2x^2+10x)+(5x+25)=0

2x(x+5)+5(x+5)=0

(2x+5)(x+5)=0

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2 years ago

Read 2 more answers

A circular pond 24 yd in

Schach [20]

Answer:

The walk will cost $8164.

Step-by-step explanation:

Given:

Diameter of the circular pond (D) = 24 yd

Width of the gravel path (x) = 2 yd

Cost per yard of the path = $50

Now, radius of the circular pond is half of the diameter and is given as:

$Radius,R=\frac{D}{2}=\frac{24}{2}=12\ yd$

Now, area of the pond is given as:

$A_{pond}=\pi R^2=3.14\times (12)^2=3.14\times 144=452.16\ yd^2$

Area of the complete path including the pond area is given as:

$A_{outer}=\pi(R+x)^2=3.14\times(12+2)^2=3.14\times196=615.44\ yd^2$

Now, area of the gravel path can be obtained by subtracting the pond area from the total outer area. This gives,

$A_{path}=A_{outer}-A_{pond}\\\\A_{path}=615.44-452.16=163.28\ yd^2$

Now, using unitary method,

Cost of 1 square yard of path = $50

∴ Cost of 163.28 square yard of path = 50 × 163.28 = $8164

Hence, the walk will cost $8164.

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3 years ago

Read 2 more answers

Given the equation ay + bx - c = 0, solve for the variable c

Gre4nikov [31]

Answer:

Step-by-step explanation:

a= -bx+c divided by y

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2 years ago

This is kind of off topic for math but I still need to know the answer, so I’ve finished all my math work and I know my work is

Mariulka [41]

Answer:

email or call or talk to your teacher and ask why

Step-by-step explanation:

6 0

2 years ago

Which of the following can be used to calculate the volume of water inside the fish bowl?

Lunna [17]

The volume of a sphere can be calculated as

$V=\frac{4}{3}\pi r^3$

Where r is the radius of the sphere

We want to calculate half of the volume, then we must divide that volume by 2

$V^{\prime}=\frac{1}{2}\frac{4}{3}\pi r^3$

Now we must find the radius of our sphere, the segment AB is the diameter of the sphere, and the radius is half od the diameter, then

$r=\frac{AB}{2}=\frac{12}{2}=6$

Let's put it into our equation

$\begin{gathered} V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(r)^3 \\ \\ V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)(\pi)(6)^3 \end{gathered}$

The problem says to use

$\pi\approx\frac{22}{7}$

Then

$V^{\prime}=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

Final answer:

The formula that can be used to calculate the volume of water inside the fish bowl is

$V=\frac{1}{2}\left(\frac{4}{3}\right)\left(\frac{22}{7}\right)(6)^3$

8 0

1 year ago