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3 years ago

Write using exponents. Rewrite the expression below in the same sequence: 4·4·4·a·b·a·b

Mathematics

2 answers:

spayn [35]3 years ago

5 0

4^3 ab^2

Four would be cubed (to the third power) and ab would be squared (to the second power)

ValentinkaMS [17]3 years ago

5 0

The answer your looking for is 4^3 a^2 b^2 and in that exact order

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

3 years ago

In this exercise, we are conducting many hypothesis tests to test a claim. Assume that the null hypothesis is true. If 400 tests

Brums [2.3K]

Answer:

Since the null hypothesis is true, finding the significance is a type I error.

The probability of the year I error = level of significance = 0.05.

so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5

Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.

3 0

3 years ago

In right triangle CAB, find the hypotenuse, CB<br> A 4 root 13<br> B 20<br> C 4 root 5<br> D 208

svetoff [14.1K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Add and simplify 1/5 + 1/10

aliya0001 [1]

In order to add these fractions we have to make a common denominator.

1/5 can turn into 2/10 Its the same thing. And its easier to solve with.

now lets multiply.

$\frac{2}{10}$ × $\frac{1}{10}$

Now we add across

2 + 1 = 3

------------

10 + 10 = 10 ( When adding fractions like these the denominator stays the same)

So your answer is...

$\frac{3}{10}$

Good Luck! :)

7 0

3 years ago

Solve for the positive value of x: 6x2 – 54 = 0

n200080 [17]

If you would like to solve for the positive value of x, you can do this using the following steps:

6 * x^2 - 54 = 0 /6
x^2 - 9 = 0
(x - 3) * (x + 3) = 0
1. x = 3
2. x = - 3

The correct result would be x = 3.

4 0

3 years ago