there are 12 students including 4 boys in a classroom. Find the number of girls in 8 such classrooms

The inequality equation of the lines are given as 4x – y ≤ 4 and 2x – 2y ≤ 3. Then the correct option is A.

Inequality is defined as an equation that does not contain an equal sign. Inequality is a term that describes a statement's relative size and can be used to compare these two claims.

The graph is given below.

From the graph, equation of the lines are given as

4x - y = 4 and 2x - 2y = 3

But for the inequality is shown in graph. Then the inequality equation of the lines are given as

4x – y ≤ 4 and 2x – 2y ≤ 3

Then the correct option is A.

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Answer:

Step-by-step explanation:

Comment

Find the number of square yards

Area = L * w

L = 2

w = 1

Area = 2 * 1

Area = 2

Now for the counter top

1 sq yard costs 204.63

2 sq yards cost 2 * 204.63

2 sq yards cost 409.26

Answer: 409.26

This shows that 154mm is equivalent to 15centimeters 4millimeters

Units are measure of a quantity. From the given question, we are to convert 15cm4mm to millimeters

Since 10mm =1cm, hence;

15cm4mm = (15 * 10) + 4

15cm4mm = 150+ 4

15cm4mm = 154mm

This shows that 154mm is equivalent to1 5centimeters 4millimeters

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$ (20b + 80) is the cost to buy a frame for the panting given that measures b meters by 4 meters and charges $10 per meter for a wooden frame. This can be obtained by finding the perimeter of the painting and multiplying with the cost of wood per meter.

Given that,

length of the painting (l) = b meters

width of the painting (b) = 4 meters

Perimeter of the painting = 2(l+b)

                                          = 2(b+4)

                                          = 2b + 8

Cost of the frame = (2b + 8)× $10 per meter

                             = $ (20b + 80)

Hence $ (20b + 80) is the cost to buy a frame for the panting given that measures b meters by 4 meters and charges $10 per meter for a wooden frame.

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x+3 is not a factor of f(x)=x^4+6x^3+5x^2-15x-2 because there is a remainder of 7 when we use the long division method to divide x+3 with the given function f(x).

The factors of a quadratic equation are the simplified form of the quadratic equation, that if multiplied together, results in the original quadratic equation.

Given that:

[tex]\mathbf{=\dfrac{x^4+6x^3+5x^2-15x-2}{x+3} }[/tex]

[tex]\mathbf{=x^3+3x^2-4x-3+\dfrac{7}{x+3} }[/tex]

x+3 is not a factor of f(x)=x^4+6x^3+5x^2-15x-2 because there is a remainder of 7 when we use the long division method to divide x+3 with the given function f(x).

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Answer:

60 in²

Step-by-step explanation:

Area of a Regular Polygon

  [tex]\textsf{A}=\dfrac{n\:s\:a}{2}[/tex]

where:

From inspection of the diagram:

Substitute the values into the formula and solve for A:

[tex]\implies \sf A = \dfrac{5 \times 6 \times 4}{2}[/tex]

[tex]\implies \sf A = \dfrac{120}{2}[/tex]

[tex]\implies \sf A = 60\:in^2[/tex]

Therefore, the area of the given regular pentagon IJKLM is 60 in².

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{First thing, let's get the formula for}\\\huge\textbf{the regular pentagon/polygon.}[/tex]

[tex]\mathbf{\dfrac{1}{2}\times \boxed{a}pothem\times\boxed{p}erimeter = {\boxed{a}rea}}[/tex]

[tex]\huge\textbf{Or we could simply say....}[/tex]

[tex]\mathbf{\dfrac{\boxed{n}umber\ of\ sides\ you\ have \times \boxed{s}ide \ length\ on\ one\ side\times \boxed{a}pothem}{2} = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Let's plug the pieces to the equation,}\\\huge\textbf{since we have that information out the}\\\huge\textbf{way.}[/tex]

[tex]\bullet\mathbf{\ Number\ of\ sides \rightarrow \boxed{\bf 5}}[/tex]

[tex]\bullet\mathbf{\ Side\ length\ of\ one\ side \rightarrow \boxed{\bf 6\ inches}}[/tex]

[tex]\bullet\mathbf{\ Apothem \rightarrow \boxed{\bf 4\ inches}}[/tex]

[tex]\huge\textbf{Making the equation for you:}[/tex]

[tex]\mathbf{\dfrac{5\times 6 \times 4}{2} = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Solving for the result:}[/tex]

[tex]\mathbf{\dfrac{5\times 6 \times 4}{2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{30\times 4}{2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{120}{2} =\boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{120\div2}{2\div2} = \boxed{a}rea}[/tex]

[tex]\mathbf{\dfrac{60}{1} = \boxed{a}rea}[/tex]

[tex]\mathbf{60\div1 = \boxed{a}rea}[/tex]

[tex]\mathbf{60 = \boxed{a}rea}[/tex]

[tex]\huge\textbf{Met the result....}[/tex]

[tex]\huge\text{Thus, your answer should be: \boxed{\mathsf{60\ in^2}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assingment \& enjoy your day!}[/tex]

Answer:

in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. Given : ABCD is a quadrilateral in which ∠A = ∠C and ∠B = ∠D . We know that the sum of the angles of a quadrilateral is 360o. Hence, ABCD is a parallelogram.

Answer:

80 - 30

Step-by-step explanation:

I'm just guessing

Answer:

c=104.1 ft

Step-by-step explanation:

Law of Cosines

c^2 = 66^2 + 120^2 - 2 (66)(120)cos 60

c = 104.1 ft

Answer:

table A

Step-by-step explanation:

A direct variation can be expressed as: [tex]\frac{y}{x}=k[/tex]. The y/x is equal to a constant value k. Multiply both sides by x you get: [tex]y=kx[/tex]. So as x increases by 1, y increases by some constant value. So to find which one will create a direct variation, see which of the tables has a constant y/x

Table A:
   7/2 = 3.5

   14/4 =3.5

   17.5/5=3.5

   21/6 = 3.5

   this has a constant y/x so it is a direct variation

Table B:

   7/2 = 3.5

   14/4 = 3.5

   21 / 5 = 4.2

   this is not a direct variation, it doesn't have a constant y/x

table C:

   7 / 2 = 3.5

   14 / 4 = 3.5

   15/5 = 3

   This is not a direct variation, it doesn't have a constant y/x

table d:
   7/2 = 3.5

   14/4 =3.5

   17/5=3.4

   This is not a direction variation, it doesn't have a constant y/x

Answer:

3rd option : 5

Step-by-step explanation:

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its a slope question

f(x) is y

y2-y1/x2-x1

-10+25/-1+4

15/3

5

Since we know that the equation of a line in slope-intercept form is: , where, m= Slope and b= y-intercept.

Upon substituting m = 5 in slope-intercept form of equation we will get,

Now let us find y-intercept by substituting coordinates of one point in our given line.

So the equation of given line will be,

Now let us find value of n by substituting x = n and y = 20 in our equation.

Therefore, the value of n is 5.

Answer:

4x+22

Step-by-step explanation:

distribute 4(x+4)

it would become 4x+16+6

16+6 = 22

4x+22

The coordinates of point P along AB so that the ratio of AP to PB is 3 to 1 is (4,13).

A coordinate system in geometry is a system that employs one or more integers, or coordinates, to define the position of points or other geometric components on a manifold such as Euclidean space.

Given that the ratio of the two lines is 3:1, therefore, we can write,

m:n = 3:1

Also, the coordinate of the endpoints of the line are,

A = (x₁, y₁) = (1, 7)

B = (x₂, y₂) = (5, 15)

Now, Using the Section formula the coordinate of the point P can be written as,

[tex]P = (\dfrac{mx_2+nx_1}{m+n}\ , \dfrac{my_2+yx_1}{m+n})\\\\P = (\dfrac{3(5)+1(1)}{4}\ , \dfrac{3(15)+1(7)}{4})\\\\P = (4, 13)[/tex]

Hence, the coordinates of point P along AB so that the ratio of AP to PB is 3 to 1 is (4,13).

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The scale factor of the image under a dilation centered at the origin is 2.

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

Translation is the movement of a point either up, down, left or right on the coordinate plane.

Let us assume that quadrilateral ABCD have vertices at A(2, 1) while its image has vertices at A'(4, 2), hence:

Scale factor = 4/2 = 2/1 = 2

The scale factor of the image under a dilation centered at the origin is 2.

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[tex]x=57^{\circ}[/tex] (alternate segment theorem)

[tex]y=67^{\circ}[/tex] (alternate segment theorem)

The rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.

We assume the station to be at point A.

The plane when directly over the station, takes point C.

As it is flying horizontally continuously at an altitude of 1 mile, AC = b = 1 mile.

The plane flies continuously, and when it is at a distance of 3 miles from the station, it is at point B.

The distance between the plane and the station of 3 miles can be shown as AB = c = 3 miles.

The movement of the plane from C to B can be shown as BC = a.

Now, from the given case, we get a right triangle ABC.

By Pythagoras' Theorem,

a² + b² = c² ... (i).

or, a² + 1² = 3²,

or, a² = 9 - 1 = 8,

or, a = 2√2 miles.

We are told that the plane is moving at a speed of 570 miles/hour.

This can be shown as, da/dt = 570 miles/hour.

As the plane is moving horizontally, the vertical distance between the plane and the station doesn't change, that is, db/dt = 0.

In the question, we are asked to find the rate at which the distance from the plane to the station is increasing, that is, we are asked to find dc/dt.

Differentiating (i), with respect to time t, we get:

2a(da/dt) + 2b(db/dt) = 2c(dc/dt).

Substituting values of a = 2√2 miles, b = 1 mile, c = 3 miles, da/dt = 570 miles/hour, and db/dt = 0, we get:

2(2√2)(570) + 2(1)(0) = 2(3)(dc/dt),

or, 2280√2 + 0 = 6(dc/dt),

or, dc/dt = 380√2 miles/hour = 537.40 miles/hour ≈ 537 miles/hour.

Thus, the rate at which the distance from the plane to the station increases is 537 miles/hour. Computed using Pythagoras Theorem and Differentiation.

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The maximum volume of the cylinder is 27147.355 at the maximum point  [tex]r = \frac{50}{\sqrt{3} }[/tex] .

The formula for the volume of the cylinder v = [tex]\pi[/tex][tex]r^{2}[/tex]h, To find the maximum volume of the cylinder we apply the condition of maxima  [tex]\frac{\mathrm{d} v}{\mathrm{d} r}[/tex] = 0.

Let r cm be the radius and h cm be the height of the closed cylinder.

Then, Total surface area of the cylinder = [tex]2\pi r(r+h)[/tex]

                                                        [tex]5000[/tex] = [tex]2\pi r^{2} +2\pi rh[/tex]

                                                             [tex]h[/tex] =   [tex]\frac{5000-2\pi r^{2} }{2\pi rh}[/tex] .............(1)

Volume of the cylinder v = [tex]\pi r^{2} h[/tex]

Substitute the value of [tex]h[/tex] in the above equation

[tex]\Rightarrow[/tex]                                       = [tex]\pi r^{2}[/tex] × [tex]\left [ \frac{5000-2\pi r^{2}}{2\pi r} \right ][/tex]

[tex]\Rightarrow[/tex]                                      = [tex]\frac{r}{2}[/tex] ×  [tex](5000-2\pi r^{2} )[/tex]

[tex]\Rightarrow[/tex]                                    v = [tex]2500r-\pi r^{3}[/tex] ..............(2)

Now, for the maximum volume of the cylinder  [tex]\frac{\mathrm{d} v}{\mathrm{d} r}[/tex] = 0

[tex]\Rightarrow[/tex]                                                            [tex]\frac{\mathrm{d} (2500r-\pi r^{3})}{\mathrm{d} r} = 0[/tex]

[tex]\Rightarrow[/tex]                                                                      [tex]3\pi r^{2} = 2500[/tex]

[tex]\Rightarrow[/tex]                                                                          [tex]r^{2} = \frac{2500}{3\pi }[/tex]

[tex]\Rightarrow[/tex]                                                                          [tex]r = \frac{50}{\sqrt{3\pi } }[/tex]

Volume is maximum for  [tex]r = \frac{50}{\sqrt{3\pi } }[/tex]  

Then, v = [tex]2500r-\pi r^{3}[/tex]

[tex]\Rightarrow[/tex]              =  [tex]2500 \frac{50}{\sqrt{3\pi } } -\pi (\frac{50}{\sqrt{3\pi } } )^{3}[/tex]                              

[tex]\Rightarrow[/tex]             = [tex]\frac{125000}{\sqrt{3\pi } } - \frac{125000}{3\sqrt{3\pi } }[/tex]

[tex]\Rightarrow[/tex]             = [tex]\frac{125000}{\sqrt{3\pi } }[/tex]×[tex]\frac{2}{3}[/tex]

[tex]\Rightarrow[/tex]             = [tex]\frac{250000}{3\sqrt{3\pi } }[/tex]                                                              

[tex]\Rightarrow[/tex]          v = 27147.355

Hence, The maximum volume of the cylinder is 27147.355 at the maximum point  [tex]r = \frac{50}{\sqrt{3} }[/tex] .                                                

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Answer:

D

21√2/2

C

C

C

Step-by-step explanation:

For all problems, the ratios of the lengths of the sides are:

30-60-90 triangle 1 : √3 : 2

45-45-90 triangle 1 : √2 : √2

Problem 3:

30-60-90 triangle at left

Hypotenuse = 11

Short leg = 11/2

Long leg = 11/2 × √3

45-45-90 triangle at right

x = 11/2 × √3 × √2 = 11√6/2

Answer: D

Problem 4:

30-60-90 triangle at top

Hypotenuse = 7√3

Short leg = 7√3/2

Long leg = 7√3/2 × √3 = 7 × 3/2 = 21/2

45-45-90 triangle at right

x = 21/2 × √2 = 21√2/2

Answer: 21√2/2

Problem 7:

30-60-90 triangle

Long leg = 4√3

Short leg = y = 4√2/2 = 2√2

Answer: C.

Problem 6:

30-60-90 triangle at left

Hypotenuse = 12√6

Short leg = 12√6/2 = 6√6

Long leg = 6√6 × √3 = 6√2√3√3 = 18√2

45-45-90 triangle at right

x = 18√2/√2 = 18

Answer: C.

Problem 8:

45-45-90 triangle

Hypotenuse = 2

Leg = y = 2/√2 = 2√2/(√2√2) = √2

Answer: C.

The pattern in the values as the exponents increase is multiply the previous value by 2.

Exponents and powers are ways used to represent very large numbers or very small numbers in a simplified manner.

Given:

Powers of 2      Values

[tex]2^{-5}[/tex]                         1/32

[tex]2^{-4}[/tex]                           1/16

[tex]2^{-3}[/tex]                           1/8

[tex]2^{-2}[/tex]                           1/4

[tex]2^{-1}[/tex]                           1/2

As, from the given data we can see that negative powers of 2 reducing by 1.

as, 2^-5 * 2= 2^-4

Hence, multiply the previous value by 2 to obtain the next value.

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Answer:

a.

Step-by-step explanation:

a= y-k/ (x-h)^2

The data can be shown in the frequency table with only 4 intervals and the histogram is shown in the picture.

It is defined as the depiction of numerical data in a graph using a bar with no space. The histogram depicts the approximate distribution of the data.

We have data:

6, 5, 4, 3, 0, 7, 1, 5, 4, 4, 3, 2, 2, 3, 2, 4, 3, 4, 0, 7

Intervals   Frequency

 0 - 3     6

 3 - 6     11

 6 - 9     3

 9 - 12     0

The histogram is shown in the attached picture.

Thus, the data can be shown in the frequency table with only 4 intervals and the histogram is shown in the picture.

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The confidence interval of employes individuals from home

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

A simple random sample size n = 250 . Of the 250 employed individuals ​surveyed,42 responded that they did work home a minimum of once per week.

Construct 99% confidence interval for population

For proportion : 42 / 250 = 1/10 = 0.16

Mean = 2.5 * sqrt [ 0.1 * 0.9 / 250]

= 2.5 * 0.01

= 0.47

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

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The value of k is  [tex]-h^2 -2xh -5x + 6[/tex]

Given;

x^(2)-5x+6=(x+h)^(2)+k

Expand the expression on the right side, we have

[tex]x^2 - 5x + 6 = (x+ h) (x+ h) + k[/tex]

[tex]x^2 -5x + 6 = x^2 +xh + xh + h^2 +k[/tex]

Collect like terms

[tex]x^2 -5x + 6 = x^2 + 2xh + h^2 + k[/tex]

[tex]x^2 -5x + 6 -x^2 - 2xh -h^2 = k[/tex]

We then have

[tex]-5x + 6 -2xh -h^2 = k[/tex]

[tex]k = -h^2 -2xh -5x + 6[/tex]

Thus, we have k as [tex]-h^2 -2xh -5x + 6[/tex]

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Answer:

B

Step-by-step explanation:

cable subscribers

Therefore, the measures of the angles in triangle ABC is as follows:

The measure of the angles in the right triangle can be found as follows:

Using trigonometric ratios,

sin ∅ = opposite / hypotenuse

sin ∅ = 7 / 25

∅ = sin ⁻¹ 0.28

∅ = 16.2602047083

∅ = 16.3°

Hence,

m∠B = 16.3°

m∠C = 90°

m∠A = 180 - 90 - 16.3 = 73.7°

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Answer:

[tex]\frac{\sqrt{3} }{3} or \frac{1}{\sqrt{3} }[/tex]

Step-by-step explanation:

We can break the problem into three parts,

tan45 = sin(45)/cos(45)=[tex]\frac{\frac{\sqrt{2}}{2} }{\frac{\sqrt{2}}{2} } =1[/tex]

[tex]tan^{2}30 =(tan30))^{2}=(sin30/cos30)^2=((1/2)/(\sqrt{3}/2))^2 =(1/\sqrt{3})^2[/tex]

[tex]tan60=sin60/cos60=\frac{\sqrt{3}/2 }{1/2} =\sqrt{3}[/tex]

Then, [tex]tan45*tan^230*tan60=1*(1/sqrt(3))^2*sqrt(3)=1/sqrt(3)=sqrt(3)/3[/tex]

Therefore, the solution is sqrt(3)/3

Answer:

7.76

Step-by-step explanation:

0.22×20=4.4

3.36(base amount ,average of the others)+4.4=7.76

The sum of the first ten prime numbers is 129.

The prime numbers are those numbers that have only one factor namely 1 and the number itself.

The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.

Their sum is

2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29

= 129

Hence, the sum of the first ten prime numbers is 129.

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