What is the equivalent expression for (3x²y÷x³)4

Jade and chet get a weekly allowance plus x dollars for each time the pair walks the dog. they plan to save 40% of their combine

irina1246 [14]

The amount of money Jade and chet will have to purchase the app is $11.6

<h3>How to solve equation?</h3>

Jade = 8 + 2x
Chet = 4x + 6

Where,

x = $2.50

Amount each earned:

Jade = 8 + 2x

= 8 + 2(2.50)

= 8 + 5

= $13

Chet = 4x + 6

= 4(2.50) + 6

= 10 + 6

= $16

Combined earnings = $13 + $16

= $29

Amount planned to save = 40% of combined earnings

= 40% × $29

= 0.4 × 29

= $11.6

Learn more about equation:

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8 0

2 years ago

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

Hitman42 [59]

Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

The solution of the intregral of equation (1) is:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

I hope it helps you!

3 0

4 years ago

When sara got to the store they sold water in gallon containers how many gallons should she buy?

sashaice [31]

Answer:

It depends how much water she would like to buy!

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, it all depends how much water she would like to buy! ^u^

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only need one more before I advance, thanks!

5 0

3 years ago

Triangle ABC is similar to triangle FGH.

Ivan

Answer:

i can't understand the question but if you put a link I can figure it out

8 0

3 years ago

In parallelogram ABCD, diagonals AC------- and BD------- intersect at point E, AE = x + 16, and CE=5x

Nataly [62]

Answer:

AC = 40

Step-by-step explanation:

A quadrilateral is a polygon shape with four sides and four angles. The interior angle of a quadrilateral sums up to 360°.

A parallelogram is a quadrilateral (has four sides and four angles) in which has two pair of opposite sides are parallel to each other. The diagonals of a parallelogram bisect each other.

Given parallelogram ABCD:

Diagonals AC and BD bisect each other at point E. Hence:

AC = AE + CE (line segment addition postulate)

Also:

AE = CE (diagonal BD bisects AC at point E).

Hence; x + 16 = 5x

5x - x = 16

4x = 16

x = 4

AE = x + 16 = 4 + 16 = 20; CE= 5x = 5(4) = 20

AC = AE + CE = 20+ 20 = 40

3 0

3 years ago