Which out of the following are expressions with numbers only?
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p

Answer:

10

Step-by-step explanation:

30-25=5

10-5=5

30:10

25:5

The graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin.

In mathematics, a graph is a visual representation or diagram that shows facts or values in an ordered way.

The relationships between two or more items are frequently represented by the points on a graph.

The y = 5 graph is a line that runs parallel to the x-axis at a distance of 5 units from the origin.

Assuming Y-axis as vertical line, X-axis as horizontal line

(see it in attached image)

Graph of y = 5 is the line where y is not changing or y always be 5, only x will change.

Graph of y is the straight line passing through 5 and parallel to X axis

Therefore, a line that is 5 units from the origin and parallel to the x-axis forms the graph of y=5.

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To graph four lines, you will need to plot at least four points for each line and then draw a line through these points.

A general procedure you can use is as follows:

It can be helpful to use a different color or style (such as a solid line or a dotted line) for each line to make it easier to tell them apart.

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The 95% confidence interval for the difference in the two means = (-2.56 , 0.16)

What is Confidence Interval?

The mean of your estimate plus and minus the range of that estimate constitutes a confidence interval. Within a specific level of confidence, this is the range of values you anticipate your estimate to fall within if you repeat the test. In statistics, confidence is another word for probability.

Given,

Sample statistics                         size(n)               mean(x)                s.d(s)

Construction site                           16                  x₁ = 5.584               1.812

Undistributed Location                 16                  x₂ = 6.789               1.945

Standard error = 0.665

Degree of freedom = 30

Critical T-value for 95% confidence interval is 2.0423

The 95% confidence interval for the difference in the two means

(construction site - undistributed location)

= x₁ -x₂ ± (t-value) (standard error)

= 5.584 - 6.786 ± (2.0423) (0.665)

= -1.2020 ± 1.358130

= (-1.2020 - 1.358130 , -1.2020 + 1.358130)

= (-2.56 , 0.156)

= (-2.56 , 0.16)

for complete question see the attachment

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Given,

Sample statistics                         size(n)               mean(x)                s.d(s)

Construction site                           16                  x₁ = 5.584               1.812

Undistributed Location                 16                  x₂ = 6.789               1.945

Standard error = 0.665

Degree of freedom = 30

Critical T-value for 95% confidence interval is 2.0423

The 95% confidence interval for the difference in the two means

(construction site - undistributed location)

= x₁ -x₂ ± (t-value) (standard error)

= 5.584 - 6.786 ± (2.0423) (0.665)

= -1.2020 ± 1.358130

= (-1.2020 - 1.358130 , -1.2020 + 1.358130)

= (-2.56 , 0.156)

= (-2.56 , 0.16)

Answer:

The Answer is 12 g because 12* g =12g

Answer:

Quartic function

Step-by-step explanation:

Answer: Quartic function

Step-by-step explanation:

676000 is the required total number of different identification codes.

The total number of possible combinations here is computed as:

= Product of the many ways to select each of the characters

= 262*103 = 676000 is the required number of total ways here.

Note that there are 26 possible ways to select each of the first 2 letters and there are 10 ways to select each of the 3 digits here.

Identification Code is a randomly generated set of alpha-numeric symbols supplied by the Bank at the time the Agreement was executed and used as a personal security measure to identify the Customer in communications between the Bank and the Customer via the Customer Service Desk.

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Answer:

36 different ones

Step-by-step explanation:

When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six)

It can reach a length of 12 inches. The smallest frog can grow to be about 4% the size of the goliath.

12.4 millimeters = 1.24 centimeters is the answer.

To convert units of length in the metric system of measurement

The basic unit of length in the metric system is the meter. All units of length in the metric system are derived from the meter. The prefix “centi-“means one hundredth.

1 centimeter=1 one-hundredth of a meter

kilo- = 1000 1 kilometer (km) = 1000 meters (m)

hecto- = 100 1 hectometer (hm) = 100 m

deca- = 10 1 decameter (dam) = 10 m 1 meter (m) = 1 m

deci- = 0.1 1 decimeter (dm) = 0.1 m

centi- = 0.01 1 centimeter (cm) = 0.01 m

milli- = 0.001 1 millimeter (mm) = 0.001 m

In the metric system, converting length units involves moving the decimal point to the right or left. Listing the units in ascending order will show how many places to move the decimal point and in which direction.

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Answer:

[tex]5 * 85 = 425[/tex]

Step-by-step explanation:

⭐ What is the product of 2+ numbers?

To get the product of 5 and 85, multiply them!

You can multiply them using a calculator, or manually multiply them just like how I showed in the image attached:

Linear graph equations are one-degree equations. It is a straight-line equation. A linear equation with parameters a and b equal to zero has the conventional form ax + by + c = 0.

A linear equation has the following standard form: ax + by + c = 0.

Here, variables x and y and constants a, b, and c are used.

Also, a ≠ 0, and b ≠ 0.

For linear equations, the slope-intercept formula is: y=mx+b

Where m stands for the line's slope and b represents the y-intercept.

The three types of linear equations are point-slope, slope-intercept, and standard form.

When an algebraic equation is graphed, it always produces a straight line since each term has an exponent of 1 or 0. This type of equation is known as a linear equation.

Steps of drawing linear graphs are described below:

Step 1: Using the provided linear equation, determine the value of y with respect to x.

Step 2: Arrange these data in a table format.

Step 3: Create a graph using the points from the database.

Step 4: Connect the points to form a straight line.

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Positive is the leading coefficient. The parabola is hence downward. The maximum vertex form value will then be a negative four.

Let the point (h, k) be the vertex of the parabola and a be the leading coefficient.

Then the equation of the parabola will be given as,

y = a(x - h)² + k

The equation is given below.

y = − x² + 2x − 5

Convert the equation into a vertex form. Then we have

y = − x² + 2x − 5

y = − x² + 2x − 1 − 4

y = − (x² − 2x + 1) − 4

y = − (x − 1)² − 4

The leading coefficient is positive. So, the parabola is downward. Then the maximum value of vertex form will be negative 4.

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The complete question is given below.

How do you find the maximum of a quadratic function on a calculator?

y = − x² + 2x − 5?

To draw asymptotes, first determine the equation of the line of the asymptote. Then use the equation to plot points and draw the line. For vertical asymptotes, use the x-intercept and for horizontal asymptotes, use the y-intercept.

To draw asymptotes, first you need to determine the equation of the line of the asymptote. The equation of the asymptote is usually in the form of y = ax + b or x = ay + b. The a and b coefficients allow you to determine the slope and intercept of the line. Once you have the equation, use it to plot points and draw the line. For vertical asymptotes, use the x-intercept and for horizontal asymptotes, use the y-intercept. Then, connect the points with a smooth line. Make sure the line does not intercept the graph at any point as this would not be an asymptote. Once you have drawn the line, check to make sure it is indeed an asymptote by comparing it to the given equation. If it matches, you have successfully created an asymptote!

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Answer: 5.66

Step-by-step explanation:

1.5(x-5)= 1

1.5x - 7.5 = 1

Add 7.5 on both sides

1.5x = 8.5

Divided 1.5 into both sides

x = 5.66

The sum of all natural numbers from 1 to 100 is 5050. The total number of natural numbers in this range is 100. So, by applying this value in the formula: S = n/2[2a + (n − 1) × d], we get S=5050.

In Java, finding the sum of two or more numbers is very easy. First, declare and initialize two variables to be added. Another variable to store the sum of numbers. Apply mathematical operator (+) between the declared variable and store the result. So you simply make this: sum=sum+num; for the cycle. For example sum is 0, then you add 5 and it becomes sum=0+5 , then you add 6 and it becomes sum = 5 + 6 and so on.

Thus, the sum of all natural numbers 1 to 10 can be calculated using the formula, S= n/2[2a + (n − 1) × d], where, a is the first term, d is the difference between the two consecutive terms, and n is the total number of natural numbers from 1 to 10. Therefore, the sum of the first ten natural numbers is 55.

public class T35{

  public static void main(String[] args) {

       int nmb;

               for(nmb= 1; nmb<= 100; nmb++){

                     System.out.println(nmb);  

       }      

  }

}

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The derivative for [tex]y=eu(x)[/tex]; [tex]u(x)[/tex] is a function in terms of [tex]x[/tex] is [tex]\frac{dy}{dx}[/tex]=[tex]eu'(x)[/tex]+ [tex]u(x)e[/tex].

Definition of derivative-A derivative is the rate of change of a function with respect to a certain variable . Derivatives are fundamental to the solution of problems in calculus and differential equations.

There are certain rules of differentiation which help us to evaluate the derivatives of some particular functions. :

Power Rule.

Sum and Difference Rule.

Product Rule.

Quotient Rule.

Chain Rule.

Given: [tex]y=eu(x)[/tex]

Explanation:-This equation can be solved using the product rule of derivatives .

According to the product rule derivative of uv will be taken as -

                                            u(v)' + v(u)'

where (') represents derivative of the variable.

Therefore accordingly, [tex]y=eu(x)[/tex]

Differentiating with respect to [tex]x[/tex],

[tex]\frac{dy}{dx} =e(u(x))'+u(x)(e')[/tex]

[tex]\frac{dy}{dx} =eu'(x)+u(x)e[/tex]

So, finally the derivative of the function in terms of x is [tex]\frac{dy}{dx}[/tex] = [tex]eu'(x)+[/tex][tex]u(x)e[/tex]

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The value of p in the equation p - 6 = -14 is -8.

An equation is the statement that illustrates the variables given. In this case, two or more components are taken into consideration to describe the scenario.

In this situation, the difference between a number p and 6 is -14.

It should be noted that the equation van be illustrated as:

p - 6 = -14

Collect like terms

p = -14 + 6

p = -8

The value of p is -8.

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Yes it is divisible by 3. A number is fully divisible by three if its digit sum is also divisible by three, according to the rule of divisibility for three.

The divisibility rule is a concise and practical approach to check, often by looking at the integer's digits, whether a given integer is divisible by a given set divisor without actually executing division. Without actually doing the division procedure, you may quickly discover if a given number can be divided by a defined divisor using the divisibility test. When dividing two numbers exactly, the quotient must be an integer and the remainder must be zero.

here  

add all the numbers,

1 + 2+ 3 + 4+ 5 = 15

and 15/3 = 5

so it is divisible by 3

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Algebra is advised for ninth grade, and pre-Algebra is taught in eighth grade, according to the Common Core standards that California enacted two years ago and is just now starting to apply.

In algebra, definite or fixed numbers with values like 2, 9, 0.068, etc. are used. Introduction to algebra, advanced algebra, abstract algebra, linear algebra, and commutative algebra are only a few of the many sub-disciplines of algebra.

here ,

The option to take it in the eighth grade may be given to students who demonstrated extraordinary potential in the subject.

Algebra is often taught in the eighth grade in many schools today, while certain students who meet a certain threshold can finish the subject in the lower grades.

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Correct Question:-

Solve the following differential equation using Laplace transformation.

[tex]\rm{x''(t)+x'(t)=2,\quad x(0)=3,\quad x'(0)=1}[/tex]

where [tex]\rm{x'(t)=\dfrac{dx}{dt},\quad x''(t)=\dfrac{d^2x}{dt^2}.}[/tex]

[tex]\quad[/tex]

Solution:-

Let,

[tex]\cal{L}\{\rm{x(t)}\}=X(s)[/tex]

[tex]\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=x(t)[/tex]

Given,

[tex]\longrightarrow\rm{x''(t)+x'(t)=2}[/tex]

We take Laplace transformation of both sides of the equation.

[tex]\longrightarrow\cal{L}\{\rm{x''(t)+x'(t)}\}=\cal{L}\{\rm{2}\}[/tex]

[tex]\longrightarrow\cal{L}\{\rm{x''(t)}\}+\cal{L}\{\rm{x'(t)}\}=\cal{L}\{\rm{2}\}[/tex]

We have,

[tex]\cal{L}\{\rm{x''(t)}\}=\rm{s^2\,X(s)-s\,x(0)-x'(0)}[/tex]

[tex]\cal{L}\{\rm{x'(t)}\}=\rm{s\,X(s)-\,x(0)}[/tex]

[tex]\cal{L}\{\rm{2}\}=\rm{\dfrac{2}{s}}[/tex]

Then,

[tex]\small\text{$\longrightarrow\rm{\big[s^2\,X(s)-s\,x(0)-x'(0)\big]+\big[s\,X(s)-x(0)\big]=\dfrac{2}{s}}$}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)-(s+1)\,x(0)-x'(0)=\dfrac{2}{s}}[/tex]

Given that [tex]\rm{x(0)=3}[/tex] and [tex]\rm{x'(0)=1.}[/tex] Then,

[tex]\longrightarrow\rm{s(s+1)\,X(s)-3(s+1)-1=\dfrac{2}{s}}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)-3s-4=\dfrac{2}{s}}[/tex]

[tex]\longrightarrow\rm{s(s+1)\,X(s)=\dfrac{2}{s}+3s+4}[/tex]

[tex]\longrightarrow\rm{X(s)=\dfrac{2}{s^2(s+1)}+\dfrac{3s+4}{s(s+1)}}[/tex]

[tex]\longrightarrow\rm{X(s)=\dfrac{3s^2+4s+2}{s^2(s+1)}\quad\dots(1)}[/tex]

We will factorise this expression by partial fractions.

Assume,

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{As+B}{s^2}+\dfrac{C}{s+1}}[/tex]

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{(As+B)(s+1)+Cs^2}{s^2(s+1)}}[/tex]

[tex]\longrightarrow\rm{3s^2+4s+2=(As+B)(s+1)+Cs^2}[/tex]

[tex]\longrightarrow\rm{3s^2+4s+2=(A+C)s^2+(A+B)s+B}[/tex]

Equating corresponding coefficients,

[tex]\rm{A+C=3}[/tex]

[tex]\rm{A+B=4}[/tex]

[tex]\rm{B=2}[/tex]

Solving each equation we get,

[tex]\rm{A=2}[/tex]

[tex]\rm{B=2}[/tex]

[tex]\rm{C=1}[/tex]

Therefore,

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{2s+2}{s^2}+\dfrac{1}{s+1}}[/tex]

[tex]\longrightarrow\rm{\dfrac{3s^2+4s+2}{s^2(s+1)}=\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}[/tex]

Then (1) becomes,

[tex]\longrightarrow\rm{X(s)=\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}[/tex]

Now we will inverse Laplace transformation to obtain the solution.

[tex]\longrightarrow\cal{L}^{\rm{-1}}\{\rm{X(s)}\}=\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{2}{s}+\dfrac{2}{s^2}+\dfrac{1}{s+1}}\right\}[/tex]

[tex]\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s^2}}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s+1}}\right\}$}[/tex]

[tex]\small\text{$\longrightarrow\rm{x(t)}=\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}+\rm{2}\,\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1!}{s^{1+1}}}\right\}+\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s-(-1)}}\right\}$}[/tex]

We have,

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s}}\right\}=\rm{1}[/tex]

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1!}{s^{1+1}}}\right\}=\rm{t}[/tex]

[tex]\cal{L}^{\rm{-1}}\left\{\rm{\dfrac{1}{s-(-1)}}\right\}=\rm{e^{-t}=\dfrac{1}{e^t}}[/tex]

Hence,

[tex]\longrightarrow\rm{\underline{\underline{x(t)=2+2t+\dfrac{1}{e^t}}}}[/tex]

This is the solution to our differential equation.

[tex]\quad[/tex]

Some results of Laplace Transformation:-

[tex]\boxed{\begin{array}{c|c}&\\\rm{x(t)}=\cal{L}^{\rm{-1}}\{\rm{X(s)}\}&\rm{X(s)}=\cal{L}\{\rm{x(t)}\}\\\\=============&==================\\\\\rm{\dfrac{d^nx}{dt^n}}&\rm{s^n\,X(s)}-\displaystyle\sum_{\rm{r=0}}^{\rm{n-1}}\rm{s^{n-r-1}\,\dfrac{d^rx}{dt^r}(t=0)}\\\\----------&--------------\\\\\rm{x'(t)}&\rm{s\,X(s)-x(0)}\\\\----------&--------------\\\\\rm{x''(t)}&\rm{s^2\,X(s)-s\,x(0)-x'(0)}\\\\----------&--------------\\\\\rm{a}&\rm{\dfrac{a}{s}}\\\\----------&--------------\\\\\rm{t^n,\ n}\in\mathbb{N}&\rm{\dfrac{n!}{s^{n+1}}}\\\\----------&--------------\\\\\rm{e^{at}}&\rm{\dfrac{1}{s-a}}\\&\end{array}}[/tex]

Answer: the range is 4

Step-by-step explanation:

Answer:

Step-by-step explanation:

You want to identify the true statements regarding angles at a transversal crossing parallel lines, and perpendicular bisectors.

The Corresponding Angles theorem tells you that corresponding angles are congruent where a transversal crosses parallel lines. Since vertical angles are congruent, and angles congruent to the same angle are congruent to each other, this also means that alternate interior angles are congruent.

A bisector of a segment passes through its midpoint, a point that is equidistant from the end points. When the bisector is perpendicular to the segment, all points on the perpendicular bisector are equidistant from the segment's endpoints.

Answer:

A line.

Step-by-step explanation:

Graphing y =  -3  will give you a horizontal line 3 units below the x-axis.

To imagine this, we can create a bunch of points, each with a y coordinate of -3. For example: (-2, -3), (-1, -3), (3, -3), (6,-3). Graphing these will lead to a horizontal line!

Using (D) the binomial formula, which is used to determine probability when independent events are known, we will get the probability in the scenario.

In an experiment with two possible outcomes, the likelihood of exactly x successes on n repeated trials is known as the binomial probability (commonly called a binomial experiment).

The binomial probability is nCx⋅px⋅(1−p)n−x if the likelihood of success on a single trial is p.

When a process is repeated a certain number of times (for example, in a set of patients), the result for each patient can either be a success or a failure, the binomial distribution model enables us to calculate the probability of witnessing a defined number of "successes."

So, in the given situation we will find the probability with help of the binomial formula which is used to calculate the probability when independent events are given.

Therefore, using (D) the binomial formula, which is used to determine probability when independent events are known, we will get the probability in the scenario.

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Complete question:

We want to determine the probability of obtaining at most 4 successful operations in 10 independent surgical operations where the probability of success is the same for each operation. The appropriate formula to be used is

(A) the mean of the binomial distribution.

(B) the hypergeometric formula.

(C) the mean of the hypergeometric distribution.

(D) the binomial formula.

Answer:

(24, 41 )

Step-by-step explanation:

y = 3x - 31 → (1)

y = - 2x + 89 → (2)

substitute y = 3x - 31 into (2)

3x - 31 = - 2x + 89 ( add 2x to both sides )

5x - 31 = 89 ( add 31 to both sides )

5x = 120 ( divide both sides by 5 )

x = 24

substitute x = 24 into either of the 2 equations and solve for y

substituting into (1)

y = 3(24) - 31 = 72 - 31 = 41

solution is (24, 41 )

Answer:

(24,41)

Step-by-step explanation:

Set the 2 equations equal to each other.

3x - 31 = -2x + 89  Add 2x to both sides of the equal sign

3x + 2x -31 = -2x + 2x + 89

5x -31 = 89  Add 31 to both sides of the equal sign

5x -31 + 31 = 89 + 31

5x = 120  Divide both sides by 5

[tex]\frac{5x}{5}[/tex] = [tex]\frac{120}{5}[/tex]

x = 24  This is your x value.

Substitute 24 for x into either of the two original equations to solve for y.

y = 3x -31

y = 3(24) - 31

y = 72 - 31

y = 41  This is your y value.

Check:

y = 3x -31

41 = 3(24) - 31

41 = 72 -31

41 = 41 Checks

y = -2x + 89

41 = -2(24) + 89

41 = -48 + 89

41 = 41 Checks.

The Range={0, 1, 2, 3, 4, 5}

Domain={0, 2, 4, 6, 8, 10}

So, options A and E are correct.

Its arguments are a group of values that can be plugged into a function's domain. This set includes the x values for a function like f(x). Its range refers to the set of digits that the function considers. The function prints these values in the order they are entered after the insertion of an x value.

Another approach to defining the scope and range of functions is through the use of graphs. Since "domain" refers to the set of possible input values, the full set of input values shown on the x-axis is the domain of a graph. On the y-axis, the range's varied output values are visible.

From the graph the points are (0, 0), (1, 2), (2,4), (3,6), (4,8), (5,10)

So,  Range={0, 1, 2, 3, 4, 5}

Domain={0, 2, 4, 6, 8, 10}

Therefore, options A and E are correct.

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The linear equation 2x + 5y = 7 has  infinitely many solutions.

In this question we need to determine the number of solutions of the linear equation 2x + 5y = 7.

We know that an equation of the form ax + by + c = 0, where a, b and c are real numbers such that a ≠ 0 and b ≠ 0, is called as a linear equation in two variables.

Also as we know a linear equation in two variables has infinitely many solutions. The graph of linear equation in two variables is a straight line. Every point on the straight line represents a solution of the linear equation.

Here we have a linear equation in two variables 2x + 5y = 7.

By rearranging given linear equation we get,

5y = 2x - 7

y = (2x - 7)/5

Here we will get different values of y for different values of x

Therefore, the linear equation has infinitely many solutions.

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