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lara31 [8.8K]
3 years ago
15

Simplify 36ab divided by 40ab​

Mathematics
1 answer:
Andru [333]3 years ago
8 0
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What is 36/40 Simplified? - 9/10 is the simplified fraction for 36/40.
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Jasmine created a scale drawing of her room. She used a scale of 1 in:3 feet. The drawing is 8 inches long by 6 inches wide. Wha
sergiy2304 [10]

Answer:

The area of the room is 432 square feet

Step-by-step explanation:

Let us use the ratio method to solve the question

∵ The scale drawing of the room is 1 inch: 3 feet

→ That means each 1 inch in the drawing represents 3 feet in real

∵ The drawing is 8 inches long by 6 inches wide

∴ The drawing length of the room = 8 inches

∴ The drawing width of the room = 6 inches

→ By using the ratio method

→  inches  :  feet

→  1            :  3

→  8           :  L

→  6           :  W

→ By using the cross multiplication

∵ 1 × L = 8 × 3

∴ L = 24

∴ The actual length of the room is 24 feet

∵ 1 × W = 6 × 3

∴ W = 18

∴ The actual width of the room is 18 feet

Use the formula of the area of the rectangle to find the area of the room

∵ Area of rectangle = length × width

∴ Area of the room = 24 × 18

∴ Area of the room = 432 feet²

∴ The area of the room is 432 square feet

8 0
3 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
3 years ago
Prove Euler's identity using Euler's formula.<br> e^ix = cos x + i sin x
Korolek [52]

First list all the terms out.

e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...

Then, we can expand them.

e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...

Then, we can use the rules of raising i to a power.

e^ix = 1 + ix - x^2/2! - ix^3/3!...

Then, we can sort all the real and imaginary terms.

e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)

We can simplify this.

e^ix = cos x + i sin x

This is Euler's Formula.

What happens if we put in pi?

x = pi

e^i*pi = cos(pi) + i sin(pi)

cos(pi) = -1

i sin(pi) = 0

e^i*pi = -1 OR e^i*pi + 1 = 0

That is Euler's identity.

3 0
3 years ago
Helppppp! what is 2+6????
yaroslaw [1]

Answer:

8...I want these points lol

Step-by-step explanation:

3 0
3 years ago
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Each day in Bio 18, Derek leaves a stack of paper packets near the entry door for students to obtain. Eighty-one percent of stud
Rainbow [258]

Answer:

Step-by-step explanation:

This is a binomial probability distribution because there are only 2 possible outcomes. It is either a randomly selected student grabs a packet before being seated or the student sits first before grabbing a packet. The probability of success, p in this scenario would be that a randomly selected student sits first before grabbing a packet. Therefore,

p = 1 - 0.81 = 0.91

n = 9 students

x = number of success = 3

The probability that exactly two students sit first before grabbing a packet, P(x = 2) would be determined from the binomial probability distribution calculator. Therefore,

P(x = 2) = 0.297

5 0
3 years ago
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