Need help ASAP will reward

The actual interval of existence for the solution is all real numbers, including t = 0.

a. The characteristic equation for the differential equation is

2t dy/dt = 8y,

which gives us

dy/dt = 4/t y.

Separating variables, we have

dy/y = 4/t dt,

which integrates to

ln|y| = 4 ln|t| + C,

where C is an arbitrary constant of integration.

Taking the exponential of both sides, we obtain

|y| = Ce^(4ln|t|),

which gives us

y = C t^4.

Using the initial condition y(-1) = 2, we can find the value of C:

y(-1) = C(-1)^4

2 = C(-1)^4

C = -2.

So the solution to the initial value problem is

y(t) = -2 t^4.

b. The largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution is t ≠ 0, since the coefficient of the differential equation is undefined at t = 0.

c. The actual interval of existence for the solution is all real numbers, including t = 0.

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a) The percentage of test takers that scored 1420 or higher is of: 2.87%.

b) The percentage of test takers that scored 800 or lower is of: 10.03%.

c) The percentage of students between 1500 and 1600 is given as follows: 0.8%.

d) The number of students eligible for the honors program is given as follows: 1040 students.

The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation for this problem are given as follows:

[tex]\mu = 1050, \sigma = 195[/tex]

The proportion of scores of 1420 or higher is one subtracted by the p-value of Z when X = 1420, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (1420 - 1050)/195

Z = 1.9

Z = 1.9 has a p-value of 0.9713.

1 - 0.9713 = 0.0287 = 2.87%.

The proportion of test takers that scored 800 or lower is the p-value of  Z when X = 800, hence:

Z = (800 - 1050)/195

Z = -1.28.

Z = -1.28 has a p-value of 0.1003.

The proportion between 1500 and 1600 is the p-value of Z when X = 1600 subtracted by the p-value of Z when X = 1500, hence:

Z = (1600 - 1050)/195

Z = 2.82

Z = 2.82 has a p-value of 0.9976.

Z = (1500 - 1050)/195

Z = 2.31

Z = 2.31 has a p-value of 0.9896.

Hence:

0.9976 - 0.9896 = 0.008 = 0.8%.

The proportion above 1550 is one subtracted by the p-value of Z when X = 1550, hence:

Z = (1550 - 1050)/195

Z = 2.56

Z = 2.56 has a p-value of 0.9948.

Hence:

1 - 0.9948 = 0.0052.

Out of 200,000 students, the number is given as follows:

0.0052 x 200,000 = 1040 students.

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The factor of the expression could be [tex](x + 1)(x + 2)((x^2 -x + 1)[/tex]

Factorization is expressing a mathematical quantity in terms of multiples of smaller units of similar quantities.

factorization is the method of breaking a number into smaller numbers that multiplied together will give that original form.

We are given a expression as;

[tex]x^4 + 2x^3 + x + 2[/tex]

Thus,

[tex](x^4 + 2x^3) + (x + 2)[/tex]

Factor out x^3

[tex]x^3(x + 2) + (x + 2)[/tex]

Then;

[tex](x^3 + 1)(x + 2)[/tex]

Therefore, the factor of the given expression is;

[tex](x + 1)(x + 2)(x^2 -x + 1)[/tex]

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Answer:

Step-by-step explanation:

x-2=-9

Add 2 on both sides

x=-7

Answer:

[tex]x=11[/tex]

Step-by-step explanation:

x−2=9

Add 2 to both sides.

x=9+2

Add 9 and 2 to get 11.

x=11

pls tell me if wrong

Answer:

below

Step-by-step explanation:

This is the same as    (-1)(-2y-4u+2)   = 2y + 4u - 2

A=1/2bh and b=2A/h so

b=2A/h =2(12cm^2)/3cm=8cm

In order to find the base

We apply the area of a triangle formula to find the base

We know the formula to find the area of a triangle is,

[tex]A =\frac{1}{2}XbXh[/tex]

Here the area and the height is given

So the base is

[tex]12 = \frac{1}{2} X bX3[/tex]

Substituting the values we get

[tex]\frac{12X2}{3} = b[/tex]

[tex]8cm = b[/tex]

The perimeter of the rectangle is (6x - 2)ft and the area of the rectangle is (2x^2 - 5x - 12) square ft.

What is a rectangle?

A rectangle is a sort of quadrilateral with parallel sides that are equal to one another and four vertices that are all 90 degrees apart. Because of this, it is also known as an equiangular quadrilateral.

We are given that the length of a rectangle is (2x + 3) ft. and the width is

(x - 4) ft.

We know that perimeter of rectangle = 2(Length + Width)

So,

⇒Perimeter = 2(2x + 3 + x - 4)

⇒Perimeter = 2(3x - 1)

⇒Perimeter = (6x - 2)ft

We know that area of rectangle = Length * Width

So,

⇒Area = (2x + 3) * (x - 4)

⇒Area = 2x^2 - 8x + 3x - 12

⇒Area = (2x^2 - 5x - 12) square ft

Hence, the perimeter of the rectangle is (6x - 2)ft and the area of the rectangle is (2x^2 - 5x - 12) square ft.

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Answer:

4.

a.) -7/6 (x + 4)(x - 3) = y

b.) -6/121 (x-8)^2 = y

c.) 1/6 (x-sqrt 7)(x + sqrt 7) = y

5.

a.) -13/36 (x+2)^2 + 5 = y

Step-by-step explanation:

4. The intercept form of a quadratic equation is

[tex]y=a(x-p)(x-q)[/tex], where a is a number determining things like if the parabola opens up or down and how wide or narrow the parabola, while p and q are the x-intercepts.

Because we're given the x-intercepts and a point on the parabola, we can simply plug everything in and solve for a to find the equation of the parabola:

a.):

[tex]y=a(x-(-4))(x-3)\\y=a(x+4)(x-3)\\\\7=a(2+4)(2-3)\\7=a(6)(-1)\\7=-6a\\-7/6=a\\\\y=-7/6(x+4)(x-3)[/tex]

b.):

[tex]y=a(x-8)^2\\\\-6=a(-3-8)^2\\-6=a(-11)^2\\-6=121a\\-6/121=a\\\\y=-6/121(x-8)^2[/tex]

c.):

[tex]y=a(x-\sqrt{7})(x-(-\sqrt{7}))\\ y=a(x-\sqrt{7})(x+\sqrt{7})\\ \\ 3=a(-5-\sqrt{7})(-5+\sqrt{7})\\ 3=a(25-5\sqrt{7}+5\sqrt{7}-7)\\ 3=a(25-7)\\ 3=18a\\ 3/18=1/6=a\\ \\ 1/6(x-\sqrt{7})(x+\sqrt{7})=y[/tex]

5. The vertex form of a quadratic equation is

[tex]y=a(x-h)^2+k[/tex], where a is the same type of number as in the intercept form and (h, k) is the vertex.

Like the intercept form of the equation, we're given everything except a and thus we can plug in what we have and solve for a:

[tex]y=a(x-(-2))^2+5\\y=a(x+2)^2+5\\\\-8=a(4+2)^2+5\\-8=a(6)^2+5\\-8=36a+5\\-13=36a\\-13/36=a\\\\y=-13/36(x+2)^2+5[/tex]

Answer:

Point A: y = 552

Point B: y = 123

Step-by-step explanation:

y = -13x - 46

A: x = -46

y = -13(-46) - 46  ==> plugin -46 for x in y = -13x - 46

y = 598 - 46

y = 552

A: x = -13

y = -13(-13) - 46  ==> plugin -13 for x in y = -13x - 46

y = 169 - 46

y = 123

The frequency distribution table is given as

Length (in)                      Frequency

2.720  - 2.729                         1

2.730  - 2.739                          5

2.740  - 2.749                          9

2.750 - 2.759                         13

2.760  - 2.769                         2

Given that the sample of the 30 screw lengths

2.753 2.762 2.752 2.735 2.733 2.747 2.751 2.723 2.754 2.751 2.754 2,742 2.748 2.737 2.751 2.752 2.764 2,744 2.744 2,748 2.741 2.762 2.758 2.756 2.747 2.736 2.752 2.736 2.759 2.746

For constructing a frequency distribution :

Lower class limit = 2.720​ in.

Class width = 0.010 in

Frequency Distribution  Table :

Length (in)                      Frequency

2.720  - 2.729                         1

2.730  - 2.739                          5

2.740  - 2.749                          9

2.750 - 2.759                         13

2.760  - 2.769                         2

Total                                        30

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Using their concepts, we have that the mean will be increased and the median will remain constant.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations.

The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile, which is the middle value of the distribution.

The largest value is the 7th observation, therefore it will be included in the calculation of the mean.

Since this observation have an increased value, the sum of all observations will be increased while the number of observations remains constant,

So the mean will be increased.

Hence, for a distribution of 7 observations, we have that:

The first three observations are the first half.

The 4th observation is the median.

The last three observations are the second half.

Here only the last observation changes, not the fourth, hence the median remains constant.

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Set A is a subset of set B if every element of A is also an element of B

If every element of set A is also an element of set B, then set A is a subset of set B. This is expressed in symbols as A B or. For example, suppose you’re representing all of the nations in the globe, and set A represents Finland and Greece, while set B represents all of Europe.

If a set A is a subset of a larger set B, the circle representing set A is drawn within the circle representing set B.

If sets A and B share certain components, we can symbolize them by drawing two overlapping circles.

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The required meaning of the word sped is A special education student.

The noun is defined as any word used to identify any of class of people, places, or things are called noun.

Here,
Sped is a noun word that tends is meaning toward education, which has meaning as special education student. The plural of sped is speds while in british english sped is used as a verb past tense and past participle of speed.

Thus, the required meaning of the word sped is A special education student.

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The price of one currency in respect to another is referred to as its exchange rate.

An exchange rate in the world of finance is the price at which one currency will be exchanged for another. The majority of the time, currencies are national ones, but they can also be supra-national ones like the euro or sub-national ones like Hong Kong.

A currency's price in relation to another is known as its exchange rate. You can have fixed or fluctuating exchange rates. While a country's central bank controls fixed exchange rates, the process of supply and demand on the open market controls floating exchange rates.

The rate at which one currency can be exchanged for another between countries or economic zones is known as the exchange rate. It is crucial in assessing trade and capital movement dynamics and is used to calculate the value of various currencies in respect to one another.

Convert the currency:

£ 400 × R 19.29/£2 = R 7,716

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The coordinates (4.0 , 4.5) to the center is not the distance of radius

From the circle formula centered at origin: x² + y² = r²

Where r is the radius of the circle. and by substitution for the various values of x, y and r we have

x² + y² = 6² = 36

Also, substituting the values of the lengths of the line we have

(4)² + (4.5)² = 16 + 20.25 = 36.25 ≠ 36

This proves the diagonals of the rhombus ABCD bisect each other

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Nominal categorical data is a categorical variable with two or more values, with no order. The examples of nominal categorical data here are the issue type, country, city.

Categorical data can be divided as two as per the variables. They are nominal and ordinal. In nominal categorical data, there will be two or more data points, but they are not ranked. The hair color can be red/blond/brown/black etc. But these cannot be ranked by taking one over the other.

In ordinal categorical data, there we can assign an order on the data points. Like grading system in marks of students, size comparison like large, medium, small. Here we could rank these in order.

Year can sometimes be nominal and sometimes ordinal. It is as per the context. If we are using it to calculate time it is ordinal.

So here we could not rank of variables in country, issue type, city. So all are nominal categorical data.

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The expression that is equivalent to (16x^4)^1/2 is (64x^6)1/3.

Equivalent expressions are expressions that accomplish the same thing even when they have distinct appearances. Two algebraic expressions that are equivalent share a similar value whenever we enter exact same thing for the variable.

Given that;

[tex](16x^4)^{\frac{1}{2}}[/tex]

[tex]=\sqrt{16x^4}[/tex]

= 4x²

From the given options, (64x^6)^1/3 verifies the identity and shows that it is equivalent to (16x^4)^1/2.

[tex](16x^4)^{\frac{1}{2}} = (64x^6)^{\frac{1}{3}}[/tex]

[tex](16x^4)^{\frac{1}{2}} = (4^3)^{\frac{1}{3}} (x^6)^{\frac{1}{3}}[/tex]

[tex]4x^2= (4) (x^6)^{\frac{1}{3}}[/tex]

[tex]4x^2=4x^2[/tex]

Therefore, we can conclude that the expression that is equivalent to (16x^4)^1/2 is (64x^6)1/3

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3) The components of composition between two functions are [tex]f(n) = n^{2}+ 5^{n}[/tex] and g(n) = n² + 1.

4) The composition between functions g(n) = √(4 + n + √n) and h(n) = n⁴ + 8 is the function (g ° h)(n) = √[4 + n⁴ + 8 + √(n⁴ + 8)].

In this problem we find two cases of composition between functions, that is, a operation between two functions, so that the input of the first function (f(x)) is substituted by the entire second function (g(x)), that is:

f ° g(x) = f [g(x)]

Case 3 - If we know that [tex](f\,\circ \,g) (n) = (n^{2}+1)^{2}+5^{n^{2}+1}[/tex], then the components of the composition of functions are:

[tex]f(n) = n^{2}+ 5^{n}[/tex]

g(n) = n² + 1

Case 4 - If we know that g(n) = √(4 + n + √n) and h(n) = n⁴ + 8, then the composition between the two functions is:

(g ° h)(n) = √[4 + n⁴ + 8 + √(n⁴ + 8)]

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Answer:

The expression simplifies to:

[tex](\frac{3p^2 + 49}{42p})^2 -(\frac{3p^2 - 49}{42p})^2[/tex]

Using the difference of squares identity, the expression further simplifies to:

[tex]\frac{98p}{42p^2} = \frac{98}{42p}[/tex]

Step-by-step explanation:

The output column, S, of the given input-output table was created from the student ID 21444368 using an algorithm where if the student ID digit is an even integer, a 0 is placed in the corresponding row of column S, and if the student ID digit is an odd integer, a 1 is placed in the corresponding row of column S.

Student ID: 81725225

P Q R S

1 1 1 0

1 1 0 1

1 0 1 1

0 1 0 0

0 0 1 1

0 1 1 0

0 0 0 1

Student ID: 81725225

P: The first digit of my student ID is 8 so it is an even integer, therefore I place a 0 in row 1 of column P.

Q: The second digit of my student ID is 1 so it is an odd integer, therefore I place a 1 in row 2 of column Q.

R: The third digit of my student ID is 7 so it is an odd integer, therefore I place a 1 in row 3 of column R.

S: The fourth digit of my student ID is 2 so it is an even integer, therefore I place a 0 in row 4 of column S.

The resulting table is:

P Q R S

1 1 1 0

1 1 0 1

1 0 1 1

0 1 0 0

0 0 1 1

0 1 1 0

0 0 0 1

The output column, S, of the given input-output table was created from the student ID 21444368 using an algorithm where if the student ID digit is an even integer, a 0 is placed in the corresponding row of column S, and if the student ID digit is an odd integer, a 1 is placed in the corresponding row of column S. The same algorithm was applied to my student ID 81725225, resulting in the table shown above.

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The slope of the secant line is 137.

The rate of change of discharges between 1994 and 1995 can be determined by taking the first derivative of the function, f'(x) = 3.3x² - 72x + 260. When x = 1994, the rate of change is -6. This means that the discharges decreased by 6 from 1994 to 1995.

The average rate of change of discharges from 1994 to 2000 can be determined by taking the average of the first derivative of the function for x = 1994 and x = 2000. When x = 1994, the first derivative is -6 and when x = 2000, the first derivative is 148. The average rate of change of discharges from 1994 to 2000 is therefore 71.

f(0) = 1.2 (0)3-36 (0)2+262/0)+570 = 570

+(4) = 1-2(4)3-36(4)2 +262 (4) +570 = 1118.8

Slope of the secant line = [tex]\dfrac{1118.8 - 570}{4-0}[/tex]

Slope of the secant line = 137

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Sorry but i can't really read the writing so... Maybe try take a pic again

Answer: Oh well how you doing?

Step-by-step explanation:

If needed call prevention hotline... if you know what I mean

Answer:

that’s brazy

Step-by-step explanation:

Part A: The range of a function is the set of all possible outputs (y-values) for the given domain of the function. In this case, the domain of f(x) is all real numbers. We can find the range of f(x) by analyzing the two different parts of the piecewise function:

For x <= 2, f(x) = 2^(x+1) - 3. The minimum possible output for this function is y = -1 (when x = -1) and the maximum possible output is y = 3 (when x = 0).

For x > 2, f(x) = 10/x. The minimum possible output for this function is y = 0 (as x approaches infinity) and the maximum possible output is y = 10 (when x = 2)

So, the range of f(x) is [-1, 10].

Part B: The asymptotes of a function are the lines that the graph of the function approaches but never touches. In this case, the graph of f(x) has a vertical asymptote at x = 0, because the function is not defined at x=0.

Part C: The end behavior of a function is the behavior of the function as the input (x) approaches positive or negative infinity. In this case, the function f(x) = 10/x as x > 2, as x approaches positive infinity the output(y) approaches 0, and as x approaches negative infinity the output(y) does not exist. Therefore, the end behavior of f(x) is that it approaches 0 as x approaches positive infinity.

   The amount that should be recorded for the asset purchase is $375,000. This is the amount of the loan that is being used to purchase the asset.

   The amount that should be recorded for the liability used to purchase the asset is also $375,000. This is the amount of the loan that is being taken on to purchase the asset.

(a) Verifying that y(x) = (x - c)^2 satisfies the differential equation y' = 2y:

To verify that the function y(x) = (x - c)^2 satisfies the differential equation y' = 2y, we can find the derivative y' and substitute it back into the equation:

y' = d/dx [(x - c)^2] = 2(x - c)

So, y' = 2y if and only if 2(x - c) = 2y, which can be simplified to x - c = y.

Therefore, y(x) = (x - c)^2 satisfies the differential equation y' = 2y for all x, including x = c.

(b) Values of b for which the initial value problem y' = 2/y, y(0) = b has no solution or a unique solution:

(i) No solution: If b = 0, the initial value y(0) = 0 and the denominator of the right-hand side of the differential equation y' = 2/y would be 0, making the equation undefined at that point. Therefore, the initial value problem y' = 2/y, y(0) = 0 has no solution.

(ii) Unique solution defined for all x: For all other values of b (b ≠ 0), the initial value problem y' = 2/y, y(0) = b has a unique solution that can be found using standard methods of solving initial value problems. This solution will be defined for all x.

The figure illustrating the fact that the initial value problem y' = 2/7, y(0) = 0 has infinitely many different solutions would show a family of curves that are all solutions to the differential equation y' = 2/7, with different starting values y(0) leading to different curves. This demonstrates that there are infinitely many different solutions to the initial value problem.

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