Select the correct answer.
A school conducts 27 tests in 36 weeks. Assume the school conducts tests at a constant rate. What is the slope of the line that represents the
number of tests on the y-axis and the time in weeks on the x axis

A baseball stadium has 45,000 seats. If tickets have been bought for 5/8 of the seats for the next game, then 45000−(5/8⋅45000)=16875 seats left.

We will subtract (5/8⋅45000) from 45000.

To subtract in mathematics is to take something away from a group or a number of things. The group's total number of items decreases or becomes lower when we subtract from it. The components of a subtraction problem are the minuend, subtrahend, and difference.

Subtraction is therefore the antithesis of addition. Thus, addition and subtraction are the polar opposites of each other. We could say that subtraction is the opposite of addition.

A subtraction number sentence is an equation that displays the result of subtracting one number from another. The first number is the minuend, the second number is the subtrahend, and the answer is the difference in a subtraction number sentence.

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Answer:

32

Step-by-step explanation:

1-295/2

Answer:

x>22/3 or x<-32/3

Step-by-step explanation:

Step 1 - add 6 to both side

Step 2 - divide both sides by two

Step 3 - Solve Absolute Value

We know either x+ 5/3 >9 or x+5/3<-9

Subtract 5/3 from both sides

Answer: x∈[-4,3)     y∈(-5,5)

Step-by-step explanation:

[tex]Domain:x\in[-4,3)\\Range\ of\ y:\ y\in(-5,5)[/tex]

Charles is correct as the solution of the mathematical operation (6 + 6) × 6 is also 72 as 12 × 6 = 72.

So, 12 × 6 = 72

Now, let's do what Charles say as follows:
Break 12 as 6 + 6.

Therefore, Charles is correct as the solution of mathematical operation (6 + 6) × 6 is also 72 as 12 × 6 = 72.

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Answer: DG and EF are parallel. DE=13

Step-by-step explanation:

DG and EF are parallel. They are both right angled triangles.
EF/DG=DE/CD

5/1.25=DE/3.25

4=DE/3.25

4*3.25=DE

13=DE

Answer:

Step-by-step explanation:

The constant of variation k for the direct variation in 4x = –6y will be 1.5.

It should be noted that a direct variation simply means the mathematical relationship that exists between two variables which can be expressed by an equation such that a variable is equal to another one.

Therefore, the constant of variation k for the direct variation in 4x = –6y will be:

4x = –6y

Divide through by 4

4x / 4 = -6y / 4

x = -1.5y

Therefore, the constant of variation is 1.5.

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If [tex]x[/tex] is positive, then [tex]\tan^{-1}(x)[/tex] is between 0 and π/2, which means [tex]\cos(\tan^{-1}(x))[/tex] is between 1 and 0.

Let [tex]\theta = \tan^{-1}(x)[/tex]. Then [tex]\tan(\theta) = x[/tex], and it follows from the Pythagorean identity that

[tex]\sin^2(\theta) + \cos^2(\theta) = 1 \implies \tan^2(\theta) + 1 = \sec^2(\theta) \\\\ ~~~~ \implies \sec(\theta) = \sqrt{1 + \tan^2(\theta)} \\\\ ~~~~ \implies \cos(\theta) = \dfrac1{\sqrt{1+\tan^2(\theta)}} \\\\ ~~~~ \implies \cos(\tan^{-1}(x)) = \boxed{\dfrac1{\sqrt{1+x^2}}}[/tex]

We took the positive square root when solving for [tex]\sec(\theta)[/tex] because we know [tex]\cos(\theta)[/tex] is non-negative, between 0 and 1.

Macro bought 175 pens

The number of pens Macro bought can be calculated as follows

let x represent the number of pens bought

let y represent the number of pencils bought

x + y = 300.........equation 1

0.25x + 0.15y= 62.50.........equation 2

From equation 1

x + y= 300

x= 300-y

Substitute 300-y for x in equation 2

0.25(300-y) + 0.15y= 62.50

75 - 0.25y + 0.15y= 62.50

75 - 0.1y= 62.50

-0.1y= 62.50-75

0.1y= 12.5

y= 12.5/0.1

y= 125

Substitute 125 for y in equation 1

x + y= 300

x + 125= 300

x= 300-125

x= 175

Hence Marco bought 175 pens

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According to the given statement

Quadratic equation for x2+8x−5=0 is x=-4± √21

Second-degree algebraic expressions of the type ax2 + bx + c = 0 are known as quadratic equations.

There are a maximum of two solutions for x in the second-degree quadratic equations. These two solutions for x are denoted as (, ) and are also known as the roots of the quadratic equations.

According to the given value;

For this equation: a=1, b=8, c=-5

1x2+8x+−5=0

Step 1: Use quadratic formula with a=1, b=8, c=-5.

x=

−b±√b2−4ac

2a

x=

−(8)±√(8)2−4(1)(−5)

2(1)

x=

−8±√84

2

x=−4+√21 or x=−4−√21

Hence,quadratic equation for x2+8x−5=0 is x=-4± √21

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Answer:

m = 2

Step-by-step explanation:

Find LCM.

[tex]7,\:14\\14=7\cdot \:2\\=2\cdot \:7\\=7\cdot \:2\\=14\\[/tex]

Multiply

[tex]\frac{2m}{7}\cdot \:14+\frac{3m}{14}\cdot \:14=1\cdot \:14[/tex]

Simplify(AA)

[tex]\frac{2m}{7}\cdot \:14\\=\frac{2m\cdot \:14}{7}\\=\frac{28m}{7}\\=4m[/tex]

Simplify(AB)

[tex]\frac{3m}{14}\cdot \:14\\=\frac{3m\cdot \:14}{14}\\=3m[/tex]

Simplify(AC)

[tex]1\cdot \:14\\= 14[/tex]

Divide

[tex]=7m = 14\\\frac{7m}{7}=\boxed{\frac{14}{7}}\\= 2[/tex]

Answer:

8a^5 - 28a^4 + 2a^3 +29a^2 + 6a - 9

Step-by-step explanation:

(4a3-5a-3) × (2a2-7a+3)

= 4a^3 × 2a^2 + 4a^3(-7a) + 4a^3 × 3 - 5a × 2a^2 - 5a(-7a) - 5a × 3 - 3 × 2a^2 - 3(-7a) - 3 × 3

= 8a^5 - 28a^4 + 2a^3 +29a^2 + 6a - 9

Answer:

a diamond ♦ it as acute angles

Using proportions, the person who earned the most commission is given by:

B. Bridget.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.

The commission is given by:

Rate(as a decimal) x Number of Items x Cost of Each Item.

Hence, for Bridget, the commission is given by:

B = 0.045 x 18 x 50 = $40.5.

For Curtis, the commission is given by:

C = 0.0375 x 11 x 80 = $33.

For Gloria, the commission is given by:

G = 0.0475 x 5 x 140 = $33.25.

For Hugo, the commission is given by:

H = 0.0425 x 13 x 60 = $33.15.

Hence Bridget earned the most commission and option B is correct.

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Increase in the length is 3.64 ft.

What is area of square?

The area of square is the product of side of the square.

i.e., Area of square =  side * side

Given that;

Yossi has garage whose area =262 ft²

And, 50% area is incremented.

Now, the New Area is calculated as;

Area = 262 × 150%

       = 393 ft²

As we know that;

Area of Square = (Length) × (Length)

Length = √393 = 19.82 ft

Also,

Length of garage = √262 = 16.18 ft

Hence, Increase in the length = 19.82 - 16.18 = 3.64 ft.

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The weight of each baby based on the information will be:

Martin = 2.9 kg

Daniel = 3.5kg

Kevin = 3.9 kg

Martin = x

Daniel = x + 0.6

Kevin = x + 1

Total weight = 10.3 kg

Therefore, this will be solved as:

x + x + 0.6 + x + 1 = 10.3

3x + 1.6 = 10.3

3x = 10.3 - 1.6

3x = 8.7

Divide

x = 8.7 / 4

x = 2.9

Therefore, the weights will be:

Martin = x = 2.9 kg

Daniel = x + 0.6 = 2.9 + 0.6 = 3.5kg

Kevin = x + 1 = 2.9 + 1 = 3.9 kg

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Approximately 733395 babies were born in the UK in 2018.

Given,

In the UK in 2018 a baby was born every 43 seconds.

We have to find the total number of babies in the UK in 2018.

A baby in every 43 seconds.

There is 3600 seconds in an hour.

There is 24 hours in a day.

In 2018 there were 365 days.

So, the total number of babies born in 2018 in the UK:

= (Seconds in an hour × hours in a day × total days in 2018) ÷ 43

= (3600 × 24 × 365) ÷ 43

= 31536000 ÷ 43

= 733395.3488

≈ 733395

That is, in the UK in 2018 approximately 733395 babies were born.

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Answer:

[tex]\frac{-5}{12}[/tex]

Step-by-step explanation:

[tex]\frac{-3}{4}[/tex] - [tex]\frac{-1}{3}[/tex]  Subtracting a negative number is the same as adding a positive

[tex]\frac{-3}{4}[/tex] + [tex]\frac{1}{3}[/tex]  Rewrite each fraction with a common denominator.  Let's use 12 as our new denominator.

[tex]\frac{-3}{4}[/tex] = [tex]\frac{-9}{12}[/tex]  multiple the top and the bottom of the first fraction by 3 to get the equivalent fraction [tex]\frac{-9}{12}[/tex]

[tex]\frac{1}{3}[/tex] = [tex]\frac{4}{12}[/tex]  Multiple the top and bottom of the first fraction by 4 to get the equivalent fraction [tex]\frac{4}{12}[/tex].

Now, add the two fractions

[tex]\frac{-9}{12}[/tex] + [tex]\frac{4}{12}[/tex] = [tex]\frac{-5}{12}[/tex]  The denominator stays the same and you add the numerators.

In simplest radical form, the values are,

(1) The missing length = 48

(2) The missing length = 48

(3) The missing length = 140

(4) The missing length = 25

(1) : In triangle,

ΔADB ≅ ΔABC

∠A = ∠A

∠ABC = ∠ABD

So, ΔADB ≅ ΔABC

AD/AB = AB/AC

AD = 64 AND AC = 100 substitute in this equation,

64/AB = AB/100

AB * AB = 64*100

[tex]AB^{2}[/tex] = 6400

AB = [tex]\sqrt{6400}[/tex]

AB = 80

Because,

∠ADB = 90°

So,

[tex]AD^{2} +BD^{2} =AB^{2}[/tex]

[tex]BD^{2} = AB^{2} -AD^{2}[/tex]

BD = [tex]\sqrt{BD^{2}-AD^{2} }[/tex]

BD = [tex]\sqrt{6400-4096}[/tex]

BD = [tex]\sqrt{2304}[/tex]

BD = 48

Therefore,

The missing length is BD = 48

(2) : In triangle,

ΔADB ≅ ΔABC

∠A = ∠A

∠ABC = ∠ABD

So, ΔADB ≅ ΔABC

AD/AB = AB/AC

AD = 64 AND AC = 100 substitute in this equation,

64/AB = AB/100

AB * AB = 64*100

[tex]AB^{2}[/tex] = 6400

AB = [tex]\sqrt{6400}[/tex]

AB = 80

Because,

∠ADB = 90°

So,

[tex]AD^{2} +BD^{2} =AB^{2}[/tex]

[tex]BD^{2} = AB^{2} -AD^{2}[/tex]

BD = [tex]\sqrt{BD^{2}-AD^{2} }[/tex]

BD = [tex]\sqrt{6400-4096}[/tex]

BD = [tex]\sqrt{2304}[/tex]

BD = 48

Therefore,

The missing length is BD = 48

(3): In triangle BDC

tanθ = BC/DC

tanθ = 60/36

θ  = [tex]tan^{-1}[/tex](60/36)

θ  = [tex]tan^{-1}[/tex](1.66)

θ = 53.16

And,

[tex]BD^{2} =BC^{2} +CD^{2}[/tex]

[tex](BD)^{2}[/tex] = 60*60 + 36*36

[tex](BD)^{2}[/tex] = 3600+1296

[tex](BD)^{2}[/tex] = 4896

BD = 69.97

In big triangle, ABC

cosθ = BC/AC

cosθ = 60/(36+u)

cos69.97 = 60/(36+u)

0.342 = 60/(36+u)

36+u = 60/0.342

36+u = 195.4

u = 195.4 - 36

u = 139.4

u = 140

Therefore,

The maximum length u = 140

(4) : In ΔABC and  ΔDAB

∠B = ∠B   (Common)

∠A = ∠D = 90°

Hence, ΔBAC ≅ ΔBDA

AB/BC = BD/AB

15/x = 9/15

x = (15*15)/9

x = 225/9

x = 25

Therefore,

The missing length x = 25

Hence,

(1) The missing length = 48

(2) The missing length = 48

(3) The missing length = 140

(4) The missing length = 25

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Considering the definition of perpendicular line, the equation of of perpendicular line is y= -3/4x + 5/2.

A linear equation o line can be expressed in the form y = mx + b

where

Perpendicular lines are lines that intersect at right angles or 90° angles. If you multiply the slopes of two perpendicular lines, you get –1.

In this case, the line is 3y= 4x +7. Expressed in the form y = mx + b, you get:

y= (4x +7)÷ 3

y= 4/3x -+7/3

The line has a slope of  4/3. If you multiply the slopes of two perpendicular lines, you get –1, then:

4/3× slope perpendicular line= -1

slope perpendicular line= (-1)÷ 4/3

slope perpendicular line= -3/4

So, the perpendicular line has a form of: y= -3/4x + b

The line passes through the point (6, -2). Replacing in the expression for parallel line:

-2= (-3/4)×6 + b

-2= -9/2 + b

-2+ 9/2= b

5/2= b

Finally, the equation of of perpendicular line is y= -3/4x + 5/2.

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We have the identity

[tex](a + b)^2 = a^2 + 2ab + b^2 = a^2 + ab + b (a + b) \\\\ ~~~~ \implies a + b = \sqrt{a^2 + ab + b (a + b)} \\\\ ~~~~ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}[/tex]

assuming [tex]a+b\ge0[/tex].

For the numerator, let [tex]b=1[/tex], so [tex]a\ge-1[/tex], which means

[tex]a^2 + a = x \implies a = \dfrac{-1 + \sqrt{1 + 4x}}2[/tex]

and so

[tex]\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\cdots}}}} = \dfrac{-1 + \sqrt{1 + 4x}}2 + 1 = \dfrac{1 + \sqrt{1 + 4x}}2[/tex]

For the denominator, let [tex]b=-1[/tex] so that [tex]a\ge1[/tex]. Now

[tex]a^2 - a = x \implies a = \dfrac{1 + \sqrt{1 + 4x}}2[/tex]

so that

[tex]\sqrt{x - \sqrt{x - \sqrt{x - \sqrt{\cdots}}}} = \dfrac{1 + \sqrt{1 + 4x}}2 - 1 = \dfrac{-1 + \sqrt{1 + 4x}}2[/tex]

So, in the integral we can simplify and evaluate it to

[tex]\displaystyle \int_0^{2021} \frac{\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\cdots}}}}}{1 + \sqrt{x - \sqrt{x - \sqrt{x - \sqrt{\cdots}}}}} \, dx \\\\ ~~~~~~~~ = \int_0^{2021} \frac{\frac{1 + \sqrt{1 + 4x}}2}{1 - \frac{1 - \sqrt{1+4x}}2} \, dx \\\\ ~~~~~~~~ = \int_0^{2021} \frac{1 + \sqrt{1 + 4x}}{1 + \sqrt{1 + 4x}} \, dx \\\\ ~~~~~~~~ = \int_0^{2021} dx = \boxed{2021}[/tex]

Answer:

Choice C; degree is 3

Step-by-step explanation:

The degree of the polynomial is the greatest of the exponents (powers) of its various terms.

So the degree of this polynomial is 3  which is choice C

Answer:

-1

Step-by-step explanation:

Answer:

-1 is the answer

Step-by-step explanation:

Answer:

[tex]\sf A. \quad -3.5 < -\dfrac{10}{3}[/tex]

Step-by-step explanation:

Convert -3.5 into a fraction:

[tex]\implies \sf -3.5=-\dfrac{35}{10}=-\dfrac{\diagup\!\!\!\!5 \times 7}{\diagup\!\!\!\!5 \times 2}=-\dfrac{7}{2}[/tex]

Rewrite both fractions so that they have the same denominator:

[tex]\implies \sf -\dfrac{7}{2}=-\dfrac{7 \times 3}{2 \times 3}=-\dfrac{21}{6}[/tex]

[tex]\implies \sf -\dfrac{10}{3}=-\dfrac{10 \times 2}{3 \times 2}=-\dfrac{20}{6}[/tex]

[tex]\textsf{As }\: -\dfrac{21}{6} < -\dfrac{20}{6} \textsf{ then}:[/tex]

[tex]\sf \implies -3.5 < -\dfrac{10}{3}[/tex]

Another name for a zero sum pair is additive inverse.

A number's cumulative is its inverse number. When a number is multiplied by its cumulative , the sum of both figures equals zero. The straightforward approach is to convert a positive number to a negative number and vice versa. We formerly know that 7(- 7) Equals 0. Accordingly,-7 is the cumulative of 7, and 7 is the cumulative  of-7.

Inverse cumulative Property

When the sum of two real figures is zero, each is said to be the cumulative  of the other.

Inverse Complement of Real figures

A whole number, a natural number, an integer, a bit, a numeric, or any real number can be specified.

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Answer:

I believe its cubic polynomial because it has 3 degrees

Step-by-step explanation:

Answer:

quartic

Step-by-step explanation:

The degree of a polynomial is the highest exponent in the expression. Polynomials are classified by their degree

This polynomial has its highest exponent 4

Therefore the degree of this polynomial is 4

A polynomial with degree 4 is known as a quartic polynomial

The required solution of the given expression is -9.

A straight line with numbers placed at equal intervals or segments along its length. Number line is a straight line with a "zero" point in the middle, with positive and negative numbers listed on either side of zero and going on indefinitely.

Given:

The given expression is

-2-(+7)

According to given question we have

Number line is a line with 0 in the middle, and then increasing positive numbers to the right of the 0 and decreasing negative numbers to the left of the 0. The line is infinite in length and it's not possible to show the whole line, but we will illustrate where on the line 9 and -9 are located.

-2-(+7)

=-2-7

=-9

Therefore, the required solution of the given expression is -9.

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The algebraic expression of the mathematical statement three more than a number, multiplied by five more than the number is (3 + x) * (5 + x)

The mathematical statement is given as

Three more than a number, multiplied by five more than the number.

Let the number be represented with x.

So, we have the following illustration

Three more than a number, multiplied by five more than the number ⇒ Three more than x, multiplied by five more than x

More than means +

So, we have the following illustration

Three more than a number, multiplied by five more than the number ⇒ 3 + x, multiplied by 5 + x

Multiplied by means *

So, we have the following illustration

Three more than a number, multiplied by five more than the number ⇒ (3 + x) * (5 + x)

Hence, the algebraic expression of the mathematical statement three more than a number, multiplied by five more than the number is (3 + x) * (5 + x)

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Answer:

1/2

Step-by-step explanation:

If there are 13 men, then the ratio of men to the total is 13:26. If you simplify that, it would just be 1:2, so for every 2 students in the class, 1 of those students is a man. This would just be half.

Answer:

Step-by-step explanation: just mutiply 7x3 you get 21