HELP ASAP!!!!!!

If I bought shoes for 32.30 with 15% off 7%sales tax a 25dollar gift card but had a coupon that was forgotten for 5%off the orig

Otrada [13]

Answer:

$4.37 cash outlay

$29.37 total amount paid, partly with gift card

Step-by-step explanation:

The discount was ...

0.15 × $32.30 = $4.845 ≈ $4.85

So the price to which tax is added is ...

$32.30 -4.85 = $27.45

The amount of tax is 7% of this value, or ...

tax = $27.45 × 0.07 = $1.9215 ≈ $1.92

So, payment is made on the amount ...

$27.45 + 1.92 = $29.37

$29.37 is the total amount paid.

If $25 is paid using a gift card, then the cash outlay is $4.37.

_____

Usually only a single coupon can be applied to a given purchase. Since the 15% store discount exceeds the amount of the 5% coupon left at home, it makes more sense to claim the 15% store discount and save the 5% coupon for another time.

__

<em>Alternate interpretation</em>

If, by some magic, both discounts could be applied, then the amount to be paid is $27.65, of which $2.65 exceeds the gift card amount. The calculation proceeds the same way, except that 20% is used instead of 15%.

4 0

3 years ago

What is the end behavior of the polynomial function? PLEASE HELP!!!

konstantin123 [22]

For this case, what you should do is to see the behavior of the graph as you move away to minus infinity or to more infinity on the x axis.
We have then that:
x ---> - inf then,
y -----> - inf
Answer:
option 4

6 0

3 years ago

Read 2 more answers

In a normal distribution, which is greater, the mean or the median?

lora16 [44]

It would be mean, since it's common for the mean to be greater than the median in a normal distribution.

I hope this helps :)
Sorry if I'm wrong or anything, but I still hope I helped.

6 0

3 years ago

Read 2 more answers

Compared to its 'parent' function f(x)=x^2, describe the changes if f(x)=-x^2-3?

zimovet [89]

Answer:

it opens down meaning it has a maximum instead of a minimum. the graph has a horizontal shift of 3 spaces to the right

Step-by-step explanation:

6 0

3 years ago

Can someone explain step by step how to do this problem? Thanks! Calculus 2

Ganezh [65]

Answer:

1.314 MJ

Step-by-step explanation:

As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.

There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.

The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.

<h3>Mass of water</h3>

The mass of the water being raised is the product of the volume of the cone and the density of water.

The cone volume is ...

V = 1/3πr²h . . . . . . for radius 2 m and height 8 m

V = 1/3π(2 m)²(8 m) = 32π/3 m³

The mass of water in the cone is then ...

M = density × volume

M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg

<h3>Center of mass</h3>

The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.

The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...

W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J

As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.

V(y) = (32π/3)(y/8)³

The differential volume at height y is the derivative of this:

dV = π/16y²

The work done to raise this volume of water to a height of 10 m is ...

(9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J

The total work done is the integral over all heights:

$\displaystyle W=612.5\pi\int_0^8{(10y^2-y^3)}\,dy=\left.612.5\pi y^3\left(\dfrac{10}{3}-\dfrac{y}{4}\right)\right|_0^8\\\\W=612.5\pi\dfrac{2048}{3}\approx\boxed{1.3136\times10^6\quad\text{joules}}$

It takes about 1.31 MJ of work to empty the tank.

8 0

2 years ago