Simplifying algebraic expression

Rongopai's age is equal to the total number of candles on the cake, thus I need to determine what that is.

What I need to figure out because I am aware that Rongopai's age is equal to the total number of candles on the cake.

The number of pieces the cake was sliced into must first be determined. A quarter has a specific appearance to me. This is how it would seem (the teacher traces a quarter shape on the circle). There must have been more than four pieces since this one seems a little smaller.

Each cut's location may be identified using my finger. Given that all the pieces must be the same size, I'm predicting the locations of each line using the first piece as a reference.

Here would be the approximate location of the first line, followed by here and here (the teacher draws a line with her finger) (tracing two more lines). the third line. In my mind, the lines and parts are somewhat "visible." Four "invisible" pieces plus the one drawn in are present when I count backward.

The complete question is:

A teacher wants a student to see how visualising the size of a share (in this case, a piece of cake) can be used to work out how many shares are needed to make a whole:

“Here is a piece of Rongopai’s birthday cake. Each piece of cake has the same number of candles. How old is Rongopai?”

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Age of Rongopai is 20 years.

One fifth of a circle means if a circle can be divided into five equal parts, then one fifth is one of them.

As per the figure given, there is shown a 1 fifth of a circular birthday cake.

As all the piece of cake is equal in size so the total number of cake piece is five.

the number of candles in each piece is four.

the total number of candles present in the whole cake is = 5 × 4

the total number of candles = 20

Therefore, Rongopai is 20 years old.

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Answer: The graph is NOT [tex]\text{y} > 3\text{x}-4[/tex]

Explanation:

The inequality of [tex]\text{y} > 3\text{x}-4[/tex]  has the boundary equation [tex]\text{y} = 3\text{x}-4[/tex]

That equation is in the format [tex]\text{y} = m\text{x}+b[/tex] which is slope-intercept form.

m = 3 = slope

b = -4 = y intercept

The positive slope indicates the boundary line goes uphill when moving to the right. However, the graph shows the complete opposite with the line going downhill. This means that the graph cannot possibly be [tex]\text{y} > 3\text{x}-4[/tex]

Point P lies on the diagonal AC of square ABCD with AP > CP. Let [tex]$O_1$[/tex] and [tex]$O_2$[/tex] be the circumcenters of [tex]$\triangle \mathrm{ABP}$[/tex] and [tex]$\triangle \mathrm{CDP}$[/tex] respectively. Given that AB = 12 and [tex]$\angle \mathrm{O}_1 \mathrm{PO}_2=120^{\circ}$[/tex], then [tex]$\mathrm{AP}=\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}$[/tex],

where a and b are positive integers.

The value of a + b is 96.

Let mid-point of DC be E and mid-point of AB be F.

Since [tex]$\mathrm{O}_1$[/tex] and [tex]$\mathrm{O}_2$[/tex] are circumcenters, therefore both lie on the perpendicular bisectors of DC and AB passing through E and F.

Now, [tex]$\mathrm{O}_1 \mathrm{P}=\mathrm{O}_1 \mathrm{~B}$[/tex] and [tex]$\mathrm{O}_2 \mathrm{P}=\mathrm{O}_2 \mathrm{D}$[/tex]

Also [tex]$\angle \mathrm{CAB}=\angle \mathrm{ACD}=45^{\circ}$[/tex]

Therefore [tex]\angle \mathrm{BO}_1 \mathrm{P}=2 \times 45^{\circ}=90^{\circ}=\angle \mathrm{DO}_2 \mathrm{P}$[/tex]

Angle subtended by an arc of a circle at its center = 2 × the angle subtended at its circumference

Therefore [tex]\angle \mathrm{O}_1 \mathrm{~PB}$[/tex] and [tex]$\angle \mathrm{O}_2 \mathrm{PD}$[/tex] are isosceles right triangles.

Using the above information and symmetry,

[tex]$\angle \mathrm{DPB}=120^{\circ}$[/tex]

Also, [tex]$\triangle[/tex]ABP is congruent to [tex]$\triangle[/tex]ADP (SAS)

and [tex]$\triangle[/tex]CPB is congruent to [tex]$\triangle[/tex]CPD (SAS)

Therefore [tex]\angle \mathrm{APB}=\angle \mathrm{APD}=60^{\circ}[/tex]

And [tex]\angle \mathrm{CPB}=\angle \mathrm{CPD}=120^{\circ}[/tex]

Now, [tex]$\angle \mathrm{ABP}=(180-60-45)^{\circ}=75^{\circ}$[/tex]

[tex]\Rightarrow \angle \mathrm{O}_1 \mathrm{BF}=30^{\circ}[/tex]

And [tex]$\angle \mathrm{PDC}=(180-120-45)^{\circ}=15^{\circ}$[/tex]

[tex]\Rightarrow \angle \mathrm{O}_2 \mathrm{DE}=30^{\circ}[/tex]

Therefore [tex]\triangle \mathrm{O}_1 \mathrm{BF}$[/tex] and [tex]$\triangle \mathrm{O}_2 \mathrm{DE}$[/tex] are right angled triangles.

BF = DE = [tex]\frac{12}{2}[/tex] = 6

[tex]\Rightarrow O_1 B=O_2 D=4 \sqrt{3}[/tex]

And  PB = PD = [tex]\frac{4 \sqrt{3}}{\cos 45^{\circ}}=4 \sqrt{6}[/tex]

Now, Let x = AP

Using law of cosine on [tex]$\triangle[/tex]ABP, we have

[tex]& 96=x^2+144-24 x \times \frac{1}{\sqrt{2}} \\[/tex]

[tex]& \Rightarrow x^2-12 \sqrt{2} x+48=0 \\[/tex]

[tex]& \Rightarrow x=\sqrt{72} \pm \sqrt{24}[/tex]

Taking the positive root, AP = [tex]\sqrt{72}+\sqrt{24}$[/tex]

Therefore a + b = 72 + 24 = 96

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Point P lies on the diagonal AC of square ABCD with AP > CP. Let [tex]$O_1$[/tex] and [tex]$O_2$[/tex] be the circumcenters of [tex]$\triangle \mathrm{ABP}$[/tex] and [tex]$\triangle \mathrm{CDP}$[/tex] respectively. Given that AB = 12 and [tex]$\angle \mathrm{O}_1 \mathrm{PO}_2=120^{\circ}$[/tex], then [tex]$\mathrm{AP}=\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}$[/tex], where a and b are positive integers. Find a + b. (correct answer +5, wrong answer 0 )

A 69 sided shape called"Hexacontakaienneagon".

Naming a polygon by its number of sides requires that we know the appropriate:

3=tri, 4=tetra, 5=penta, 6=Hexa, 7=hepta, 8=octa, 9=Nona or ennea, 10=deca,11=hendeca,12=dodeca,13=triskaideca,14=tetradeca,15=pentadeca,16=hexadeca,17=heptadeca,18=octadeca,19=enneadeca,20=icosa

For polygons with 3 through 20 sides, simply add "gon" to the prefixes at the left.

For more than 20 sides, we "construct" the name by using so-called combining prefixes:

For Tens Digit:

20=icosi,30=triaconta,40=tetraconta,50=pentaconta,60=hexaconta,70=heptaconta,80=octaconta,90=enneaconta

for One's Digit:

1=hena,2=di,3=tri,4=tetra,5=penta,6=hexa,7=hepta,8=octa,9=ennea

60                and      9    

hexaconta    kai     ennea

As per the given information, 69 is defined as "Hexacontakaienneagon".

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As a result, the response to the query that the area provided With interiors of 32 and 44 square feet, respectively, the future complex will consist of a 60.1-foot circular.

Finding out how big a rectangle or rectangle is is the simplest (and most used) method. Divide the width by the altitude to find the area of a rectangle. Since all sides of a square are equal in length, determining the area of unit only requires knowledge of the size of one of its sides.

Here,

Here is the formula for area:

The area formula is length times width.

=> 8 × 4

= >32ft²

Following space will be on the trapezoid:

=>(14 + 18)/2 × 4

=> 44ft²

By multiplying the measurement by two, the perimeter will indeed be determined. Knowing this,

= 2(8 + 4) + (14 + 8 + 4 + 5.6 + 4.5)

= 24 + 36.1

= 60.1 ft

In light of this, the answer to the area-related question is

For the specified polygon, 60.1 square feet.

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Answer: 32

Step-by-step explanation: 4 x 8 = 32 . I am concerned how simple this is ;-;

Answer:

32

Step-by-step explanation:

each boxes have 4 muffins

so,in 8 boxes

by question

8×4=32

The answer is :6.5g

Inverse operations, such as addition and subtraction, multiplication and division, are pairs of mathematical operations where one operation reverses the effect of the other. The term "inverse" refers to the reciprocal of a number, i.e., x - 1 = 1 / x. One results from multiplying a number by its reciprocal, or inverse.

Inverse operations, such as addition and subtraction, multiplication and division, are pairs of mathematical operations where one operation reverses the effect of the other. The term "inverse" refers to the reciprocal of a number, i.e., x - 1 = 1 / x. One results from multiplying a number by its reciprocal, or inverse.

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The parent function of a line (usually of the form y = ax + b) is y = x.

The most basic form of a function is a parent function. Its essential shape is not changed in any manner. In mathematics, some graphs are frequently seen. Because of this, the graphs of quadratic functions, square roots, absolute values, cubics, and cube roots are among the original, widespread functions that are referred to as parent graphs.

For instance, you should still be able to identify the parabola that has undergone various transformations when you view a u-shaped graph that has been inverted and stretched vertically.

Linear functions have x as the term with the highest degree and a general form of y = a + bx. All linear functions have a straight line as a graph. The parent function of linear functions is y = x, and it passes through the origin. The domain and range of all linear functions are all real numbers.

These functions represent relationships between two objects that are linearly proportional to each other.

Therefore, the parent function of a line (usually of the form y = ax + b) is y = x.

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Here the point 3 can be taken as (3,0) or (0, 3) ; For first, move right 3 units along the x-axis. No vertical component is there , as the y-coordinate is 0 here; For second, move 3 units along the y-axis. Since the x-coordinate is 0, there is no horizontal component.

A point is marked in a coordinate plane using its coordinates when it is plotted in a coordinate system.

The intersection of the x-axis, a horizontal number line, and the y-axis, a vertical number line, creates a 2D coordinate plane.

At the coordinate plane's origin, where the two axes (plural for axis) intersect, four right angles are formed. On a coordinate plane, a grid is frequently drawn to make it simpler to plot points.

In a 2D coordinate plane, a point's location is represented by an ordered pair of numbers (x, y). An ordered pair of numbers is used to represent the location of a point in a 2D coordinate plane (x, y). The first number in the pair represents the point's location along the x-axis, and the second number represents its location vertically (y-axis).

Count x squares from the origin along the x-axis, then count y squares in the direction of the y-axis to plot the point (x, y) in the coordinate plane.

Here the point 3 can be taken as (3,0) or (0, 3) ; For first, move right 3 units along the x-axis. No vertical component is there , as the y-coordinate is 0 here; For second, move 3 units along the y-axis. Since the x-coordinate is 0, there is no horizontal component.

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The coordinates of the image of the square KLMN are:

K = (0, -1)

L = (2, -1)

M = (2, 1)

N = (0, 1)

The graph of the image is given below.

The coordinates in a graph indicate the location of a point with respect to the x-axis and y-axis.

The coordinates in a graph show the relationship between the information plotted on the given x-axis and y-axis.

We have,

The coordinates of the square KLMN are:

K = (6, -5)

L = (8, -5)

M = (8, -3)

N = (6, -3)

Now,

Translation of 6 units left and 4 units up.

This means,

K = (6 - 6, -5 + 4) = (0, -1)

L = (8 - 6, -5 + 4) = (2, -1)

M = (8 - 6, -3 + 4) = (2, 1)

N = (6 - 6, -3 + 4) = (0, 1)

Thus,

The image of the square KLMN will be on the following coordinates.

K = (0, -1)

L = (2, -1)

M = (2, 1)

N = (0, 1)

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Answer:

F.

Step-by-step explanation:

y = k/x where k is a constant

given that x = 4 when y = 3, we have:

3 = k/4

So, k =12.

The equation of variation is:

y = 12/x

When x = 15

y = 12/15 = 4/5.

So not 12 so this is False,

Answer: the cost for Internet increase by 5/4 per minute

Step-by-step explanation:

2) y+ 3÷y is the equation that is not a polynomial. Using addition, subtraction, multiplication, and division of numbers .

Using just the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables, a polynomial is an expression in mathematics made up of variables (also known as indeterminates) and coefficients. Using addition, subtraction, multiplication, and division of numbers, a polynomial is an equation that combines variables, constants, and exponents (No division operation by a variable).

given

An algebraic mathematical equation known as a polynomial might have two components. coefficients and parameters

You can combine the solutions to these algebraic equations to achieve the result.

It follows the general formula as follows:

a + a1 + a+ …

2)y+ 3y is the equation that deviates from the formula above.

So, 2)y+ 3y is the equation that is not a polynomial.

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The volume of the cone is (2/3)π cubic feet. And the radius of the spherical barrier would be 0.8 ft. and the height would be 2r = 2(0.8) = 1.6 ft. The correct option is A.

Space occupied by an object in three-dimensional space is called as volume of an object. In simple words, space is taken by object.

Given:

Tobias is constructing barriers to prevent cars from driving on a bike path. He can make conical pylons using the dimensions shown.

He used the same amount of concrete to construct a spherical barrier.

According to the information  given in the exercise, the height and the radius of the pylons are the following:

Height h = 2 ft

Radius r = 1 ft

So, the volume of the cone,

= (1/3)πr²h

= (1/3)π(1)²(2)

= (2/3)π cubic feet.

And the volume of the sphere is twice the volume of the cone.

So, the volume of the sphere,

4πr³/3 = 2π/3

2r³ = 1

r³ = 1/2 = 0.5

r = ∛0.5 = 0.8 ft.

Therefore, the radius of the spherical barrier would be 0.8 ft. and the height would be 2r = 2(0.8) = 1.6 ft.

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To find the radius of a circle given the area of a sector and the measure of the central angle of the sector, we can use the formula:

radius = sector area / (central angle measure / 360)

In this case, the sector area is 165.6 cm2 and the central angle measure is 129*. Plugging these values into the formula gives us:

radius = 165.6 / (129 / 360)

= 165.6 / (129 / 360)

= 165.6 / (0.3583)

= 462.9 cm

Rounding this value to the nearest tenth gives us a radius of approximately 463 cm.

Question 1

The area of the fairway between two streams on a golf course cannot be determined without additional details such as the dimensions or measurements of the fairway and the streams.

Question 2

A. Triangle

To find the area of the fairway, you could separate it into figures whose areas you can find using formulas.

Step 1: Draw a diagram of the fairway, including the two streams and any other relevant features such as bunkers or trees.

Step 2: Divide the fairway into two sections, one on either side of the streams.

Step 3: Measure the width and length of each section of the fairway.

Step 4: Use the formula for the area of a rectangle to find the area of each section. (Area = width x length)

Step 5: Add the areas of the two sections together to find the total area of the fairway between the two streams.

The base of the trapezoid would be the width of the fairway and the width of the stream, and the height would be the distance between the fairway and the stream. The area of a trapezoid is (a+b)h÷2, where a and b are the lengths of the bases and h is the height. The triangle would be formed by the side of the trapezoid that is the distance between the fairway and the stream, and the width of the stream. The area of a triangle is base×height÷2.

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Answer:

∠8 = 139°

Step-by-step explanation:

Before we start solving, let's identify the figure to understand the problem.

We are given two parallel lines that are cut by a transversal.

Two lines are parallel if they have the same arrow head marking.

If two parallel lines are cut by a transversal, the following angles are created:

I have attached an image of the different types of angles to this response.

Thus, we can use the types of angles to solve for ∠8.

∠3 and ∠5 are corresponding angles. Therefore, they have the same measure.

Set up the equation to solve for x:

∠3 = ∠5

9x + 14 = 14x - 1 . . . . . . substitute the values

14 = 5x - 1 . . . . . . . . subtract 9x from the LHS and RHS

15 = 5x . . . . . . . . . . add 1  to the LHS and RHS

∴ 3 = x . . . . . . . . . . . . divide the LHS and RHS by 5 to isolate x

Now that we know what x is, we can substitute it into the value of ∠3.

∠3 = 9x + 14

∠3 = 9(3) + 14 . . . . substitute the values

∠3 = 27 + 14 . . . . . compute

∴ ∠3 = 41

∠3 and ∠8 are adjacent angles. Therefore, they add up to 180°.

Set up the equation to solve for ∠8:

∠8 + ∠3 = 180

∠8 + 41 = 180 . . . . substitute the known values

∴ ∠8 = 139° . . . . . subtract 41 from the LHS and RHS to isolate ∠8

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Answer:

Step-by-step explanation:

180-54=139 so 8 equals 139

F:4+4=8

A:23+8+4=35

H:53+3+28=84

I:11+17+93=121

I:24+8+3=35

I think that is all correct

Answer:

y = 72 and x = 36

Step-by-step explanation:

So firstly, the outside angle is 108, right? And the total of all angles in a triangle is 180 degrees. So:
180 - 108 = 72
Then the other angle, angle x, is also 72 degrees. That's because they are supplementary angles.
Now that we know x = 72 degrees. To find the y angle, you add 72 and 72 together. Which is 144.
Then you do 180 - 144. Which is 36. So, x = 72 and y = 36

I hope it helped you!

Explain how the Distributive Property can be used to solve the equation 3(x+4)=36. Multiply the first term in the parentheses by 3, then add 3 to the second term in the parentheses. Then solve the equation using inverse operations.

The solution of the given system of equations is x = -7/3 and y = 7.

The given system of equations is given as,

3x + 2y = 7

y = 3x + 11

To solve this system of equations, we need to eliminate one of the variables. We can eliminate y by subtracting the second equation from the first.

3x + 2y = 7

y = 3x + 11

⇒ 3x + 2y - y = 7 - 11

⇒ 3x + 2y - 3x - 11 = -4

⇒ 2y - 11 = -4

⇒ 2y = -4 + 11

⇒ 2y = 7

Now, substitute the value of y in the first equation to find the value of x.

3x + 2(7) = 7

⇒ 3x + 14 = 7

⇒ 3x = 7 - 14

⇒ 3x = -7

⇒ x = -7/3

Therefore, the solution of the given system of equations is x = -7/3 and y = 7.

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Señor Ochoa draws the perimeter of the garden with actual length of the vertical and horizontal legs of his triangular garden 3 meters as 6cm in his initial drawing with 18.5cm third side. Then new drawing with the vertical and horizontal legs of 18 cm

A scale drawing is a representation of an object or space that is reduced or enlarged in proportion to the actual size. To make a new scale drawing of Señor Ochoa's garden with the new dimensions of 18 centimeters for the vertical and horizontal legs of the triangle, you would need to apply a scale factor of 18/6 = 3 to the original drawing.

So, you would multiply each original dimension by 3 to get the new dimensions for the scale drawing.

The new scale drawing would be:

6cm × 3 = 18cm

6cm × 3 = 18cm

8.5cm × 3 = 25.5cm

The new scale is

= scale dimension/ original dimension

= 18cm/ 3m

= 18cm/ 300cm

= 3/ 50

It is also worth noting that as you are changing the scale, the measurement of the hypotenuse of the triangle would be affected, which is the square root of 3 meters, that would be approximately 25.5cm, which is the same as the original measurement on the original drawing.

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The required expression that has the coefficient 4 is 4x⁵. Option B is correct.

The expression is the relationship between variables and represented as y = ax + b is an example of a polynomial expression.

here,
Since the question seems to be incomplete,
A standard question can be given as,
Among the following option find the expression that has a coefficient of 4.
(a) 2x + 1
(b) 4x⁵
(c) ax⁶
(d) 7x²

Since the coefficient is the value present at the prefix of the variable term,
So the variable term that holds the coefficient 4 is 4x⁵.

Thus, the required expression is 4x⁵. Option B is correct.

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Answer:

h = 4 ft

Step-by-step explanation:

the area (A) of a parallelogram is calculated as

A = bh ( b is the base and h the perpendicular height )

here b = 8 and h = 3 , then

A = 8 × 3 = 24 ft²

Now consider the other base b = 6 with perpendicular height h , then

6h = 24 ( divide both sides by 6 )

h = 4

The volume of air present in the room is 27410.7075 lt .

The space occupied inside an object's borders in three dimensions is referred to as its volume. It is sometimes referred to as the object's capacity.

Each thing in three dimensions takes up some space. The volume of this area is what is being measured. The space occupied inside an object's borders in three dimensions is referred to as its volume. It is sometimes referred to as the object's capacity.

The capacity occupied by a three-dimensional solid shape is known as volume. It is difficult to imagine in any shape, yet it may be compared amongst shapes. For instance, a compass box has a larger volume than an eraser placed inside of it. We split the area into equal square units to get the area of any two-dimensional form. Similar to this, we shall split the volume of solid objects into equal cubical units while calculating it. In the part after this one, let's examine how to determine the volume of various solid forms.

The dimension of the room is 11*11*8 in feet.

so, 11 ft = 11 *2.54*12 = 335.28 cm

And, 8 ft = 8 *2.54*12 = 243.84 cm

Volume of the room is = 335.28 * 335.28 * 243.84 = 27410707.5 [tex]cm\x^{3}[/tex]

In litre,

27410707.5 [tex]cm\x^{3}[/tex] = 27410707.5/1000 = 27410.7075 lt.

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