Let a,b,c be positive real numbers. Prove the inequality
[tex]\displaystyle\\\bf\frac{(a+b)^2}{c} +\frac{c^2}{a} \geq 4b[/tex]

An equation for the line in slope-intercept form is y=(6/5)x+8

The same number is given for every two different points on the same line when the ratio is stated as a quotient ("rise over run"). On a falling line, there is a negative "rise." In a diagram that depicts a roof or a road as a description or a design, the line may be practical as established by a road surveyor.

The steepness, incline, or grade of a line can be approximated by the slope's absolute value. The slope of a line determines how steep it is.

From the graph,

The line passes through the points (0,8) and (5, 14)

Slope m=(14-8)/(5-0)

m=6/5

(0, 8) passes through the line y=mx+c

So, 8=6(0)/5+c

Then, c=8

An equation for the line in slope-intercept form is y=(6/5)x+8.

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The coordinates point C of the given line segment AC is C(0, -2) whose midpoint is at B.

To calculate the midpoint of a line segment, find the average for the coordinates of the endpoints of the line segment. I.e.,

Consider a line segment AC, B is its midpoint.

So, the coordinates of B are (Where we have A(x1, y1) and C(x2, y2))

B(x, y) = [tex](\frac{x1+x2}{2}, \frac{y1+y2}{2})[/tex]

It is given that, B is the midpoint of a line segment AC.

Where A has coordinates as (-3, 4) and B has coordinates as (-1 1/2, 1).

So, the coordinates of the other end point C of the given line segment are calculated as

B(x, y) = [tex](\frac{x1+x2}{2}, \frac{y1+y2}{2})[/tex]

⇒ (-1.5, 1) = [tex](\frac{-3+x2}{2}, \frac{4+y2}{2})[/tex]

⇒ -1.5 = (-3 + x2)/2 and 1 = (4 + y2)/2

⇒ -3 = -3 + x2 and 2 = 4 + y2

⇒ x2 = 0 and y2 = 2 - 4 = -2

Therefore, point C has coordinates as (0, -2).

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The equivalent of the expression 2³/2⁷ using the law of indices is (a) 1/2⁴

From the question, we have the following parameters that can be used in our computation:

2³/2⁷

The above expression can be solved using the law of indices

When the law of indices is applied, we have the following representation

2³/2⁷ = 2(³ ⁻ ⁷)

Remove the bracket in the above expression

So, we have the following representation

2³/2⁷ = 2³ ⁻ ⁷

Evaluate the difference

This gives

2³/2⁷ = 2⁻⁴

Rewrite as

2³/2⁷ = 1/2⁴

Hence, the solution is (a) 1/2⁴

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The reason why looking back is important in mathematical answers is that; Option D: Looking back allows a student to check that the answer is correct and makes sense in context of the problem.

We know that in mathematics, every problem has steps to solve it whether algebra, proportion, trigonometry, calculus, congruency, angles, logarithms, e,t,c

Now, since we know that every Maths problem has steps to solve them, then we can say that, in some instances, we definitely have to look back to be sure that the answer actually makes sense when we consider the solution being required.

Now, if we have to look back to check that the answer is actually correct and properly portrays the solution being demanded, then we can say that the appropriate option for looking back her is option D.

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Answer:

The line and the parabola cross at x = -4 and x = 1

Step-by-step explanation:

Set f(x) = g(x)

[tex]x-1 = (x+2)^2-9\\x - 1 = x^2+4x+4-9\\x = x^2 + 4x - 4\\0 = x^2 + 3x - 4\\Factor\\0 = (x - 1) (x + 4)\\x = 1, -4[/tex]

Answer:

On average, the football club sells 1,058,000 tickets per season.

Step-by-step explanation:

[tex]23000[/tex] × [tex]46 = 1058000[/tex]

Answer:

Factorization

Step-by-step explanation:

Check the picture below.  Let's recall is an isosceles, so it has twin sides and is a right-triangle.

[tex]x^2+(\sqrt{6x+16})^2\implies \sqrt{x^2+6x+16}\qquad \impliedby hypotenuse \\\\[-0.35em] ~\dotfill\\\\ x~~ = ~~\sqrt{6x+16}\implies \stackrel{\textit{squaring both sides}}{x^2=6x+16}\implies x^2-6x-16=0 \\\\\\ (x-8)(x+2)=0\implies x= \begin{cases} 8 ~~ \textit{\LARGE \checkmark}\\\\ -2 ~~ \bigotimes \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\underset{\textit{length of each side}}{\stackrel{\sqrt{(8)^2 + 6(8) + 16}}{{\Large \begin{array}{llll} 8\sqrt{2} \end{array}}}\hspace{5em}\stackrel{x}{\text{\Large 8}}\hspace{5em}\stackrel{\sqrt{6(8) - 16}}{\text{\Large 8}}}[/tex]

now, let's notice, the negative value is a legitimate value, however for this case is invalid since the lengths of the triangle can't be negative.

The cost of each drink is $5.5

The cost of each candy is $5

Using the elimination method, you may either add or subtract the equations to produce an equation in one variable. To eliminate a variable, add the equations when the coefficients of one variable are in opposition, and subtract the equations when the coefficients of one variable are in equality.

According to the given information

Let number of drinks = x

Let number of candies = y

Christian bought 8 drinks and 10 candies and it costed $94

So the equation becomes

8x + 10 y = $94

Other guy bought 9 drinks and 6 candies and it costed $79.50

So the equation becomes

9x + 6y = $79.50

By applying elimination method

9x + 6y = $79.50

Dividing by 2

3x + 2y = $26.5

multiplying the above equation by 5

15x + 10y =  $132.5

8x + 10 y = $94

Subtract the following equation

7x = 38.5

x = 5.5

substituting the value of x in above one equation we get

y = 5

So the cost of each drink is $5.5

the cost of each candy is $5

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Answer:

D = the price of each drink

C = The price of each candy

One drink costs d. So 8 drinks cost 8d. Same thing with candies except a different variable. One candy costs c. Therefore 10 candies cost 10c. So the first equation here would be :

8d + 10c = 94

Same thing with the second equation but with different numbers :

9d + 6c = 79.50

If you're also looking for a solution to this answer. You use a process called the elimination method.

Let me know if you have any questions ! <3

What is parallelogram side theorem?

The parallelogram side theorem simply means that the opposite sides of a parallelogram are equal and also parallel to each other.

Proof: We have AD || BC [Opposite sides of parallelogram ABCD]

∠BAC= ∠ACD …(i) [Alternate angles with transversal AC]

Now, AB || CD [Opposite sides of parallelogram ABCD]

∠DAC = ∠ACB …(ii) [Alternate angles with transversal AC]

In ΔABC and ΔACD, we have

∠1 = ∠3 [From (i)]

∠2 = ∠4 [From (ii)]

and AC = CA [Common]

∴ ΔABC≅ ΔCDA [By ASA Congruence]

Hence Proved.

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ΔABC≅ ΔCDA By ASA Congruence, therefore m∠ BAC ≅ m∠ DCA, hence proved using parallelogram side theorem.

What is parallelogram side theorem?

The parallelogram side theorem simply means that the opposite sides of a parallelogram are equal and also parallel to each other.

Proof: We have AD || BC [Opposite sides of parallelogram ABCD]

∠BAC= ∠ACD …(i) [Alternate angles with transversal AC]

Now, AB || CD [Opposite sides of parallelogram ABCD]

∠DAC = ∠ACB …(ii) [Alternate angles with transversal AC]

In ΔABC and ΔACD, we have

∠1 = ∠3 [From (i)]

∠2 = ∠4 [From (ii)]

and AC = CA [Common]

∴ ΔABC≅ ΔCDA [By ASA Congruence]

Hence Proved.

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Answer:

-16n-41

Step-by-step explanation:

The graph that shows a dilation among the given graphs is; Graph F.

Dilation in graphs simply means each side is multiplied by the same

scale factor. What this means is that the ratio of each dilated side to their respective old sides before dilation must be the same scale factor.

We can see that the pre-dilated image has 2 units along the y - axis and 4 units across the x-axis. After dilation, it now has 4 units along the y - axis and 8 units across the x-axis. This means the scale factor in both axis are; 4/2 = 2 and 8/4 = 2. They are both the same scale factors and as such this graph shows a dilation.

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The inequality Fx + Wy > R written with x isolated to the left sides is

Inequality refers to a way of representing relationships between variables at the left hand side of equation to the variables to the right hand side of an equation.

The signs used in inequality are of 4 types namely

The given inequality in the problem has greater than sign and it is written as Fx + Wy > R

rearranging the inequality gives

Fx + Wy > R

Fx > R - Wy

divide through by F

Fx / F> R/F - Wy/F

x > (R - Wy) / F

isolating x to the left side gives  x > (R - Wy) / F

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The 99 percent confidence interval for the true mean length of the bolt is CI = (2.8712, 3.1288)

Confidence Interval is used to tell us the degree of certainty or uncertainty that is existent in a sampling method.

The general formula for confidence interval is;

CI = x' ± z(s/√n)

where;

x' is sample mean

z is z-score at confidence level

s is sample standard deviation

n is sample size

We are given;

sample size; n = 36

Sample mean; x' = 3 inches

standard deviation; s = 0.3 inches

confidence level = 99%

z at 99% CL = 2.576

Thus;

CI = 3 ± 2.576(0.3/√36)

CI = 3 ± 0.1288

CI = (2.8712, 3.1288)

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Answer:

Making a fraction is the easiest way to do this.

The x-intercepts give solutions to a problem that equals zero.

The vertical asymptotes are values x cannot be, in a fraction the denominator can not be zero.

Horizontal asymptotes are the values y approaches. This can be accomplished by making the leading term have a coefficient of 6

Step-by-step explanation:

Micro Middle store sold the least number of video game consoles.

The two dimensions of a matrix are represented by rows and columns. The row index and the column index are the two subscripts that each element is specified by.

The rows and columns in a matrix stand in for the two dimensions. The row index and the column index serve as two subscripts that each define an element. A 2D matrix is expanded into a multidimensional array, which uses additional subscripts for indexing.

The total no. of consoles sold by each store is:

Game Go = 53+21+48 = 122

Better Buy = 65+34+70 = 169

Micro Middle = 18+11+15 = 44

The least number of consoles were sold by Micro Middle(44).

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Answer:

904.32 cm³

Step-by-step explanation:

Assuming 12cm is the diameter.

The given solid is a sphere, we must find the volume.

Given radius:

[tex]6[/tex]6

Use the equation [tex]v=\frac{4}{3} \pi r^3[/tex] to find the volume of the sphere.

[tex]\frac{4}{3} \times\pi\times6^3[/tex]

≈ 904.77868

Rounding, 904.32 cm³

The required percentage is 95%  data lies between the given data.

What is percentage?

Let X be the lengths of screws.

X~ Normal (μ = 5.5, σ = 0.1)

P[5.3 < X < 5.7]

= P[(5.3-5.5)/0.1 < (X-μ)/σ < (5.7-5.5)/0.1]

= P[-2 < Z < 2] Z is a standard normal variable.

= 2*P[Z < 2] - 1

= 2*0.97725 - 1, from standard normal table.

= 0.9545

P[5.3 < X < 5.7] = 0.95

Hence, the required percentage is 95%  data lies between the given data.

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Answer:

The answer is A

Step-by-step explanation:

took the test.

Probability of winning a grand prize is 1/10

Probability of winning a basic prize is 11/20

What is the probability of two independent events?

If A and B are two independent events, then P(A and B) = P(A) * P(B).

If Events A and B are independent, the probability that either Event A or Event B occurs is: P(A or B) = P(A) + P(B) - P(A and B).

Solution:

P(A) = no. of A / total possibility

      = 4/10

P(B) = no. of 1's/total possibility

      = 3/12

     =  1/4

Now,

To win a grand prize you need an "A" and "1" .

∴ P(A and B) = P(A) × P(B)

                    = 4/10 × 1/4

         P(A and B) = 1/10

Now ,

To win a basic prize you need to spin an "A" or a "1"

∴ P(A or B) = P(A) + P(B) - P(A and B)

                  = 4/10 + 1/4 - 1/10

                  = 3/10 + 1/4

∴ P(A or B)=  11/20

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Probability of winning a grand prize is 1/10

Probability of winning a basic prize is 11/20

If A and B are two independent events, then P(A and B) = P(A) * P(B).

If Events A and B are independent, the probability that either Event A or Event B occurs is: P(A or B) = P(A) + P(B) - P(A and B).

P(A) = no. of A / total possibility

    = 4/10

P(B) = no. of 1's/total possibility

     = 3/12

    =  1/4

Now,

To win a grand prize you need an "A" and "1" .

∴ P(A and B) = P(A) × P(B)

                   = 4/10 × 1/4

        P(A and B) = 1/10

Now ,

To win a basic prize you need to spin an "A" or a "1"

∴ P(A or B) = P(A) + P(B) - P(A and B)

                 = 4/10 + 1/4 - 1/10

                 = 3/10 + 1/4

∴ P(A or B)=  11/20

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a) The probability that random guessing will lead to a passing score is of: 0%.

b) The probability that random guessing will lead to a perfect score is of: 3.35 x 10^-18.

c) The first version of the test is safer against examinees guessing to achieve a passing score.

The mass probability formula, giving the probability of x successes, is presented as follows:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are given by:

For item a, considering that for 50 questions, all are guessed among five answer, the values of these parameters are given as follows:

n = 50, p = 1/5 = 0.2.

The probability of passing is given as follows:

P(X ≥ 35).

Using a binomial distribution calculator, with the given parameters, we have that:

P(X ≥ 35) = 0%.

For item b, the parameters are given as follows:

n = 25, p = 1/4 = 0.2.

Hence the probability of a perfect score is given as follows:

P(X = 25) = 0.2^25 = 3.35 x 10^-18.

For item c, we have that the first version is safer against examinees guessing to achieve a passing score, as:

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Answer:

Hi!

The answer would be: 5 + 10m

I hope this helped you! :)

Answer:

[tex]\bf 5+10m[/tex]

Step-by-step explanation:

[tex]\bf 5\left(1+2m\right)[/tex]

Apply the distributive law:-

[tex]\bf 5\left(1+2m\right)=5\times \:1+5\times\:2m[/tex]

[tex]\bf 5\times\:1+5\times \:2m[/tex]

[tex]\bf 5+10m[/tex]

___________________

Hope  this helps!

Have a good one!

This data collection may be characterized as numerical and univariate.

If the collection of all potential values for a numerical univariate data is finite or countably infinite, the data is discrete. Usually, discrete univariate data are used in counting (such as the number of books read by a person). If the range of all potential values is a set of numbers, then a numerical univariate data is continuous.

Categorical variables represent different sorts of data that may be categorized. Race, sex, age group, and educational level are some examples of categorical variables.

Data types that may be kept and recognized based on the names or labels assigned to them are referred to as categorical data. Numerical data refers to the data that is in the form of numbers, and not in any language or descriptive form. Also known as qualitative data as it qualifies data before classifying it.

Over here we have the data for the average time spent reading per day for students in an 8th grade class.

Average Time is a numerical data

Bivariate data deals with two variables

univariate data, which is one variable data

we have numerical data and one variables so its

Numerical and univariate

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Answer:

y = 3x - 8

Step-by-step explanation:

the regular daily charge is "x", oddly enough that's the 100%.

so the new charge is 17% up, that means the new charge is 100% + 17% = 117%, and we know that's $135, so

[tex]\begin{array}{ccll} amount&\%\\ \cline{1-2} x & 100\\ 135& 117 \end{array} \implies \cfrac{x}{135}~~=~~\cfrac{100}{117} \\\\\\ 117x=13500\implies x=\cfrac{13500}{117}\implies x=\cfrac{1500}{13}\implies x=115\frac{5}{13}[/tex]

The measure of angle ABC is 90° and the the measure of angle BDC is 45°.

Square ABCD is given.

A square is a quadrilateral with four equal sides. There are many objects around us that are in the shape of a square. Each square shape is identified by its equal sides and its interior angles that are equal to 90°.

Diagonals bisects the angles in square and intersect each other at right angle.

Now, ∠ABC=90°

∠AED=90°

∠BAC=45°

∠DBC=45°

∠BDC=45°

Hence, the measure of angle ABC is 90° and the the measure of angle BDC is 45°.

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[tex] \huge{ \bold{ \color{purple}{Solution }}}[/tex]

Let a be the first term and r be the common ratio

Now,

[tex] \large{ \sf{T _3 = 4}}[/tex]

[tex] \large{ \sf{ {ar}^{2} = 4}}[/tex]

[tex]\large{ \sf{ \therefore \: product \: of \: first \: 5 \: term \: = a _1 \times a_2 \times a_3 \times a_4 \times \: a _5}}[/tex]

[tex] \large{ \sf{ = a(ar)( {ar}^{2} )( {ar}^{3} )( {ar}^{4} ) = {a}^{5} r {}^{10} }}[/tex]

[tex] \large{ \sf{ \red{(ar {}^{2} ) {}^{5} = {4}^{5} }}}[/tex]

[tex] \rule{150pt}{5pt}[/tex]