4j-2/2=4j+5

4j minus 2 divided by 2 equals 4j plus 5 ?

The expression that shows identity property of multiplication is (x+yi)×1=(x+yi), the correct option is D.

According to the identity property of multiplication, the product of 1 and a number is the number itself.

The expressions in the question are

(X+yi)×z=(xz+yzi)

(x+yi)×0=0

(x+yi)×(z+wi)=(z+wi)×(x+yi)

(x+yi)×1=(x+yi)

The only expression that shows this property is (x+yi)×1=(x+yi).

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Final answer:

The general solution to the given second-order differential equation, 3y'' + 2y' + y = 0, is found using the characteristic equation method, resulting in complex roots. The solution is expressed in terms of sine and cosine functions multiplied by an exponential decay factor.

Explanation:

To find the general solution of the given second-order differential equation, 3y'' + 2y' + y = 0, we first convert it into its characteristic equation. This is done by substituting y = ert into the differential equation, where r is the root of the characteristic equation and t is an independent variable. This approach transforms the given differential equation into a quadratic equation.

The characteristic equation for this differential equation is 3r2 + 2r + 1 = 0. Solving this quadratic equation using the formula r = [-b ± sqrt(b2 - 4ac)] / 2a, where a=3, b=2, and c=1, gives the roots of the characteristic equation. In this case, the discriminant (b2 - 4ac) is less than zero, indicating complex roots.

The roots can be found to be r = -1/3 ± i(sqrt(2)/3). Therefore, the general solution to the differential equation is y(t) = e-t/3[C1cos(sqrt(2)t/3) + C2sin(sqrt(2)t/3)], where C1 and C2 are constants determined by initial conditions.

The general solution is [tex]\( y(t) = e^{-\frac{t}{3}} (C_1 \cos\left(\frac{\sqrt{2} t}{3}\right) + C_2 \sin\left(\frac{\sqrt{2} t}{3}\right)) \)[/tex].

To solve the second-order linear homogeneous differential equation [tex]\( 3y'' + 2y' + y = 0 \)[/tex], we follow these steps:

1. Write the characteristic equation associated with the differential equation.

2. Solve the characteristic equation for its roots.

3. Write the general solution based on the roots of the characteristic equation.

Step 1: Write the Characteristic Equation

The given differential equation is:

[tex]\[ 3y'' + 2y' + y = 0 \][/tex]

We assume a solution of the form [tex]\( y = e^{rt} \)[/tex]. Substituting [tex]\( y = e^{rt} \)[/tex] into the differential equation, we get:

[tex]\[ 3(r^2 e^{rt}) + 2(r e^{rt}) + e^{rt} = 0 \][/tex]

Dividing through by [tex]\( e^{rt} \)[/tex] (which is never zero), we obtain the characteristic equation:

[tex]\[ 3r^2 + 2r + 1 = 0 \][/tex]

Step 2: Solve the Characteristic Equation

The characteristic equation is a quadratic equation:

[tex]\[ 3r^2 + 2r + 1 = 0 \][/tex]

To find the roots of this quadratic equation, we use the quadratic formula:

[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where [tex]\( a = 3 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 1 \)[/tex].

Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:

[tex]\[ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \][/tex]

[tex]\[ r = \frac{-2 \pm \sqrt{4 - 12}}{6} \][/tex]

[tex]\[ r = \frac{-2 \pm \sqrt{-8}}{6} \][/tex]

[tex]\[ r = \frac{-2 \pm 2i\sqrt{2}}{6} \][/tex]

[tex]\[ r = \frac{-1 \pm i\sqrt{2}}{3} \][/tex]

Thus, the roots of the characteristic equation are:

[tex]\[ r_1 = \frac{-1 + i\sqrt{2}}{3} \][/tex]

[tex]\[ r_2 = \frac{-1 - i\sqrt{2}}{3} \][/tex]

Step 3: Write the General Solution

Since the roots are complex conjugates [tex]\( r_1 = \alpha + i\beta \)[/tex] and [tex]\( r_2 = \alpha - i\beta \)[/tex] with [tex]\( \alpha = -\frac{1}{3} \)[/tex] and [tex]\( \beta = \frac{\sqrt{2}}{3} \)[/tex], the general solution to the differential equation is of the form:

[tex]\[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \][/tex]

Substitute [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex]:

[tex]\[ y(t) = e^{-\frac{t}{3}} \left( C_1 \cos\left( \frac{\sqrt{2} t}{3} \right) + C_2 \sin\left( \frac{\sqrt{2} t}{3} \right) \right) \][/tex]

Thus, the general solution of the differential equation [tex]\( 3y'' + 2y' + y = 0 \)[/tex] is:

[tex]\[ y(t) = e^{-\frac{t}{3}} \left( C_1 \cos\left( \frac{\sqrt{2} t}{3} \right) + C_2 \sin\left( \frac{\sqrt{2} t}{3} \right) \right) \][/tex]

where [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex] are arbitrary constants.

We can convert inches to meters with the following conversion formula:

1 inch = 0.0254 meters

Now we have to find how many meters are there in 276 inches.

So multiplying 0.0254 with 276 would give us the answer.

276 inches = 0.0254 *276 = 7.0104 meters.

Answer : There are 7.0104 meters in 276 inches.

The equivalent fractions with the same denominator are 6/ 15 and 5 /15.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that the two fractions are 2/5 and 1/3. The two equivalent expressions with the common base will be written as:-

2/5 = ( 2 x 3 ) / ( 5 x 3 )

2//5 = 6 / 15

1/3 = ( 1 x 5 ) / ( 3 x 5 )

1 / 3 = 5 / 15

Therefore, the two fractions are 6/ 15 and 5 /15.

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Answer:

The Edg answer is as follows -

The statement is not true for all real numbers. It is only true for decimal or fractional values of x .

If x is an integer, then the ceiling function returns x when the input is x . But, the flooring function returns x + 1 when the input is x + 1.

Since x does not equal x + 1 when x is an integer, the expressions are not equal when x is an integer.

Step-by-step explanation:

Real numbers are set of numbers that include rational and irrational numbers.

[x]=[x+1] is not true for all real numbers

Assume that: x is an integer

And x = 5

So, the equation becomes

[tex]5 = 5 + 1[/tex]

[tex]5 = 6[/tex]

The above equation is not true, because:

[tex]5 \ne 6[/tex]

Hence, [x]=[x+1] is not true for all real numbers

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The derivative of the function f'(x)= arccsc 8x is f'(x)= -1/(|8x|√((8x)² - 1)). We achieved this result by using the rules for derivatives of arcsine, arcsecant and arccosecant along with the chain rule for differentiation of composite functions.

To find the derivative of the function f'(x)= arccsc 8x, we will first need to understand that the derivative of the arcsine of x, also known as the inverse sine of x, is 1/(√(1 - x²)). Similarly, the derivative of arcsecant of x, known as the inverse secant of x, is 1/(|x|√(x² - 1)).

However, here we have arccosecant of x, known as the inverse cosecant of x. With this, we find that the derivative of arccosecant of x is -1/(|x|√(x² - 1)). Now in context of our function, where we replace x with 8x, we get f'(x)= -1/(|8x|√((8x)² - 1)).

We used the rules mentioned as well as the mutation rule ( df (u) = [du f(u)] dx ). This mutation rule is also known as the chain rule in differentiation, which allows us to differentiate composite functions. The function present here is a composite function where we have 8x in place of x in the arccosecant function.

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Jensen stopped at A and planned to stop at B next. He has a map with a scale of 1 inch to 35 miles.

It is clear that 1 inch on map corresponds to 35 miles in actual.

From the map he has, we can see that both places A and B are 2.5 inches apart on map.

Let's assume those are "X" miles away in actual. Then we can use proportions to solve this problem.

1 inch to 35 miles would be same as 2.5 inches to X miles. Mathematically, we can write it as follows :-

[tex] \frac{1 \;inch}{35 \;miles} =\frac{2.5 \;inches}{X \;miles} \\\\
\frac{X \;miles}{35 \;miles} =\frac{2.5 \;inches}{1 \; inch} \\\\
\frac{X}{35} =\frac{2.5}{1} \\\\
Cross \;\;multiplying \\\\
X = 35*2.5=87.5 \;miles [/tex]

Hence, actual distance between A and B is 87.5 miles.

Answer with explanation:

Question asked by Mr.Maddox to find the sum of fractions with the help of number line :

    [tex]\rightarrow 3 \frac{2}{3}+1 \frac{3}{4}+\frac{2}{3}\\\\ \text{Using Associative Property}\\\\\rightarrow a+(b+c)=(b+c)+a\\\\\rightarrow [\frac{2}{3}+ 1\frac{3}{4}]+3 \frac{2}{3}\\\\\rightarrow [\frac{2}{3}+\frac{7}{4}]+\frac{11}{3}\\\\\rightarrow \frac{21+8}{12}+ \frac{11}{3}\\\\\rightarrow\frac{29+44}{12}\\\\ \rightarrow\frac{73}{12}\\\\\rightarrow 6\frac{1}{12}[/tex]

Answer:

sample response:  None of the roots can have multiplicity because the polynomial is cubic and 3 roots are given. Write each root as a linear factor, then multiply the three factors to get the expression for the function.

Step-by-step explanation:

Final answer:

The surface area of a right rectangular prism with dimensions of 6cm by 14cm by 5cm is calculated by summing up the areas of all faces, resulting in a total surface area of 368cm².

Explanation:

To calculate the surface area of a right rectangular prism, we need to find the area of all the faces and add them together. For a prism with dimensions of 6cm by 14cm by 5cm, it has three pairs of identical faces: two faces that are 6cm by 14cm, two faces that are 6cm by 5cm, and two faces that are 14cm by 5cm.

Using the area formula for rectangles (Area = length × width), we find the areas of these faces:

For the 6cm by 14cm faces: Area = 6cm × 14cm = 84cm². Since there are two such faces, their combined area is 2 × 84cm² = 168cm².

For the 6cm by 5cm faces: Area = 6cm × 5cm = 30cm². Since there are two such faces, their combined area is 2 × 30cm² = 60cm².

For the 14cm by 5cm faces: Area = 14cm × 5cm = 70cm². Since there are two such faces, their combined area is 2 × 70cm² = 140cm².

Adding up all these areas gives us the total surface area of the prism:

Total Surface Area = 168cm² + 60cm² + 140cm² = 368cm²

Therefore, the surface area of the right rectangular prism is 368cm².

I think its both because An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant. Consecutive means one after the other. The constant value is called the common difference. Another way to think about an arithmetic sequence is that each term in the sequence is equal to the previous term plus the common difference and also A geometric sequence is a sequence in which the ratio of any two consecutive terms is constant. This constant value is called the common ratio. Another way to think about a geometric sequence is that each term is equal to the previous term times the common ratio. So when u look at it check out if the 16, 8, 4, and 2 see if they fit these terms

The ratios are constant (0.5), so the sequence is geometric.

Since the sequence is not arithmetic but is geometric, the correct answer is: B. Geometric

To determine whether the sequence 16, 8, 4, 2 is arithmetic, geometric, both, or neither, let's analyze the differences between consecutive terms and the ratios between consecutive terms.

Arithmetic Sequence:

An arithmetic sequence is a sequence in which the difference between consecutive terms is constant.

Calculating the differences:

8 - 16 = -8

4 - 8 = -4

2 - 4 = -2

The differences are not constant, so the sequence is not arithmetic.

Geometric Sequence:

A geometric sequence is a sequence in which the ratio between consecutive terms is constant.

Calculating the ratios:

8 / 16 = 0.5

4 / 8 = 0.5

2 / 4 = 0.5

The ratios are constant (0.5), so the sequence is geometric.

Since the sequence is not arithmetic but is geometric, the correct answer is:

B. Geometric

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The length in feet of the metal frame that surrounds the trampoline is 44 ft.

The circumference of a circle is the distance round the circle.

The circumference of a circle = πD

3.14 x 14 = 44 ft

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It says to check if [tex] \frac{3}{8} [/tex] satisfy each equation or not.

Checking first equation :-

96 x [tex] \frac{3}{8} [/tex]

⇒ [tex] \frac{96 * 3}{8} [/tex]

⇒ [tex] \frac{288}{8} [/tex] = 36

⇒ 36 = 36. First equation is "YES".

Checking second equation :-

38 x [tex] \frac{3}{8} [/tex]

⇒ [tex] \frac{38 * 3}{8} [/tex]

⇒ [tex] \frac{114}{8} [/tex] = [tex] \frac{57}{4} [/tex]

⇒ [tex] \frac{57}{4} [/tex] ≠ 14. Second equation is "NO".

Checking third equation :-

16 x [tex] \frac{3}{8} [/tex]

⇒ [tex] \frac{16 * 3}{8} [/tex]

⇒ [tex] \frac{48}{8} [/tex] = 6

⇒ 6 = 6. Third equation is "YES".

Checking fourth equation :-

56 x [tex] \frac{3}{8} [/tex]

⇒ [tex] \frac{56 * 3}{8} [/tex]

⇒ [tex] \frac{168}{8} [/tex] = 21

⇒ 21 = 21. Fourth equation is "YES".

Answer:  The total number of fifth graders in the school is 155.

Step-by-step explanation:  Given that at Brown Elementary School, 80% of all fifth graders ride the bus to school.

If 124 fifth graders ride the bus to school, we are to find the number of fifth graders in the the school.

Let, 'n' be the total number of fifth graders in the school.

Then, according to the given information, we have

[tex]80\%\times n=124\\\\\\\Rightarrow \dfrac{80}{100}\times n=124\\\\\\\Rightarrow \dfrac{4}{5}n=124\\\\\\\Rightarrow n=\dfrac{124\times 5}{4}\\\\\\\Rightarrow n=31\times5\\\\\Rightarrow n=155.[/tex]

Thus, the total number of fifth graders in the school is 155.

The value of x is option (D) 2

In mathematics, a function from a set X to a set Y assigns to each element of X exactly one element of Y.

Given,

f(x) = [tex]2^{x}[/tex]

we have to find the value of x when f(x) =4

f(x) = [tex]2^{x}[/tex] =4

[tex]2^{x}=4\\ 2^{x}=2^{2}[/tex]

Therefore value of x is 2

Hence, the value of x is option (D) 2

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we know that

To find out the fraction of a pan of brownies that was left subtract [tex]1\frac{1}{5}[/tex] from [tex]2[/tex]

remember that

[tex]1\frac{1}{5}=(1*5+1)/5=\frac{6}{5}[/tex]

so

[tex]2-\frac{6}{5} =(2*5-6)/5=\frac{4}{5}[/tex]

therefore

the answer is

the fraction of a pan of brownies that was left is [tex]\frac{4}{5}[/tex]

The answer is:

5 hours, 246 miles.

Since we know the equations that represent the distances of both cars at certain times (function of time), we can calculate the time that they will be at the same distance by making their equation equal.

So, we are given the equations:

A-

[tex]y=44x+26[/tex]

B -

[tex]y=36x+66[/tex]

Now, by making both equation equal, we can calculate the time that they will have the same distance, so, we have:

[tex]44x+26=36x+66[/tex]

[tex]44x-36x=66-26[/tex]

[tex]8x=40[/tex]

[tex]x=\frac{40}{8}=5[/tex]

Hence, we have that they will have the same position after 5 hours.

Then, to calculate the distance, we need to substitute the obtained time in any of the given equations, so, substituting into A, we have:

[tex]y=44x+26[/tex]

[tex]y=44*5+26=220+26=246[/tex]

We have that the distance will be 246 miles.

Hence, we have that the answer is:

5 hours, 246 miles.

Have a nice day!