All categories

3 years ago

Invests $800 into an account with a 2.4% interest rate that is compounded quarterly.

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

6 0

The equation would be 800(0.24)^x.
x would be 5.
800(1.24)^5
=2345.300
=$2345

You might be interested in

I need help finding how this would work with numerical relationships x^3-3x^2+3xy^2-y^3=(x-y)^3

diamong [38]

Answer:

y=mx+b

Step-by-step explanation:

7 0

3 years ago

What is this answer?

Karo-lina-s [1.5K]

A=7/6

since we have x^1 * x^ (1/6)

3 0

3 years ago

Read 2 more answers

Hugo is designing a frame as shown at the right. The frame has a width of x inches all the way around. Write. an expression that

Cloud [144]

The expression which shows the total area of the picture and frame is $x^{2} +72x$ +320.

Given that the width of the frame is x ,length horizontally be 20 inches and vertically be 16 inches.

We are required to find the expression which shows the total area of the picture and frame.

Expression is basically a combination of numbers, symbols, coefficients, determinants, indeterminants,etc. and are mostly not found in equal to form.

Since we are given that the width of the frame be x, So,the are of only frame will be :

Area of frame=20*x*2+16*x*2+x*x

=40x+32x+ $x^{2}$

= $x^{2}$ +72x

Area of picture=20*16=320 $inches^{2}$

The expression which shows the total area of the picture and frame is $x^{2} +72x$ +320.

Hence the expression which shows the total area of the picture and frame is $x^{2} +72x$ +320.

Learn more about expression at brainly.com/question/723406

#SPJ9

6 0

2 years ago

Could someone please help me with this problem.

lana66690 [7]

Z = 72
y = 106

hope it helps

5 0

3 years ago

Read 2 more answers

A cannonball of mass 1kg is shot vertically upward from the top of a building with an unknown velocity v_0(m/sec).v 0 (m/sec).

Bumek [7]

Taking the upward direction to be positive, the cannonball's height $y(t)$ in the air at time $t$ is given by

$y(t)=y_0+v_0 t-\dfrac g2t^2$

where $g$ is the magnitude of the acceleration due to gravity, 10 m/s^2, and $y_0$ is the height of the building from which the ball is being thrown.

At the moment the cannonball reaches its maximum height of 30 m, its velocity at that time is 0, so that

$0^2-{v_0}^2=-2g(30\,\mathrm m-y_0)\implies v_0=\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}$

Substitute this into the height equation above, and let $t=2\,\mathrm s$ , for which we have $y(2\,\mathrm s)=30\,\mathrm m$ :

$30\,\mathrm m=y_0+\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}(2\,\mathrm s)-\left(5\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)^2$