8x+2 ≥74 or 6-6x ≥30

Solve

The number of times that the shape will map back onto itself as the shape rotates 360° about its center is 6.

For a regular polygon of any number of sides, the sum of its exterior angle is 360° (full angle).

Regular polygons have all sides the same and that apothem bisects the side in two parts, (provable by symmetry).

A regular polygon with n sides can map onto itself by n times

It will rotate 360°/n about its center every time and will map onto itself

Examples are in the attached table as well as the solution.

The number of times that the shape will map back onto itself as the shape rotates 360° about its center is 6.

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Two figures are known as similar figures if there the corresponding angles are equal and the corresponding sider is in ratio. The value of x is 10. Thus, the correct option is B.

Two figures are known as similar figures if there the corresponding angles are equal and the corresponding sider is in ratio. It is denoted by the symbol "~".

Since the two prisms are similar, their sides will be in a common ratio. The value of the common ratio will be,

Common ratio = 4/2 = 5/2.5 = 3/1.5 = 2

Since the common ratio of every side is equal to 2, the value of x can be written as,

Common Ratio = 20/x = 2

20/x = 2

x = 10

Hence, the value of x is 10. Thus, the correct option is B.

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[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]

[tex] \star \: \tt \cot \theta = \dfrac{7}{8} [/tex]

[tex] {\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}[/tex]

[tex] \star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]

Consider a [tex]\triangle[/tex] ABC right angled at C and [tex]\sf \angle \: B = \theta [/tex]

Then,

‣ Base [B] = BC

‣ Perpendicular [P] = AC

‣ Hypotenuse [H] = AB

[tex] \therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}[/tex]

Let,

Base = 7k and Perpendicular = 8k, where k is any positive integer

In [tex]\triangle[/tex] ABC, H² = B² + P² by Pythagoras theorem

[tex] \longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2} [/tex]

[tex] \longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2} [/tex]

[tex]\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2} [/tex]

[tex]\longrightarrow \tt {AB}^{2} = 113{k}^{2} [/tex]

[tex]\longrightarrow \tt AB = \sqrt{113 {k}^{2} } [/tex]

[tex]\longrightarrow \tt AB = \red{ \sqrt{113} \: k}[/tex]

Calculating Sin [tex]\sf \theta [/tex]

[tex] \longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}[/tex]

[tex] \longrightarrow \tt \sin \theta = \dfrac{AC}{AB}[/tex]

[tex]\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }[/tex]

[tex]\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }[/tex]

Calculating Cos [tex]\sf \theta [/tex]

[tex] \longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}[/tex]

[tex] \longrightarrow \tt \cos \theta = \dfrac{BC}{ AB} [/tex]

[tex] \longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }[/tex]

[tex]\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }[/tex]

Solving the given expression :-

[tex] \longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } [/tex]

Putting,

• Sin [tex]\sf \theta [/tex] = [tex]\dfrac{8}{ \sqrt{113} }[/tex]

• Cos [tex]\sf \theta [/tex] = [tex]\dfrac{7}{ \sqrt{113} }[/tex]

[tex] \longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)} [/tex]

Using (a + b ) (a - b ) = a² - b²

[tex]\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} } [/tex]

[tex]\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} } [/tex]

[tex]\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} } [/tex]

[tex]\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }[/tex]

[tex]\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }[/tex]

[tex]\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64} [/tex]

[tex]\longrightarrow \: \tt \dfrac{49}{64} [/tex]

[tex]\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }[/tex]

[tex]\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered} [/tex]

[tex] {\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}[/tex]

✧ Basic Formulas of Trigonometry is given by :-

[tex]\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}[/tex]

✧ Figure in attachment

[tex]\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered} [/tex]

a. The volume of the pool is 27.483 m³

b.  The number of litres of water the pool will hold is 27483 L

c. The pool will fill at a rate of 50 L/min in 549.7 min

The pool is a cylinder, so, we need to find its volume

Since the pool is a cylinder, its volume is, V = πd²h/4 where

So, V = πd²h/4

= π(5.4 m)² × 1.20 m/4

= π × 29.16 m² × 1.20 m/4

= 34.992π m³/4

= 8.748π m³

= 27.483 m³

So, the volume of the pool is 27.483 m³

Since the volume of the pool, V = 27.483 m³ and 1000 Litres = 1 m³

So, the volume of the pool in litres is V = 27.483 m³

= 27.483 × 1 m³

= 27.483 × 1 000 L

= 27483 L

So, the number of litres of water the pool will hold is 27483 L

Since the volume of the pool is V = 27483 L, the time it will take the pool to fill at a rate of 50 L/min is given by t = Volume/rate

= 27483 L/50 L/min

= 549.66 min

≅ 549.7 min

So, how long will it take to fill the pool at a rate of 50 L/min is 549.7 min

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There will be 36 squares in the 8th step of the pattern

The pattern and the squares are represented as:

Pattern   Squares

1                 1

2                3

3               6

4             10

This can be rewritten as:

Pattern   Squares

1                 1

2                1 + 2

3                1 + 2 +3

4                1 + 2 + 3 +4

So, the number of squares on the 8th pattern is:

8th pattern = 1 + 2 + 3 +4 + 5 + 6 + 7 + 8

Evaluate

8th pattern = 36

Hence, there will be 36 squares in the 8th step

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Answer:

See below

Step-by-step explanation:

Fill the 5   and pour into the 3    ....this will leave 2 gal in the 5 bucket

  empty the 3 bucket and put that 2 gallon in to it.....

      then fill the 5 bucket again......pour 1 gallon into the 3 gallon bucket to fill it....then there will be 4 gallons left in the 5 bucket

Let

x------> the number of multiple choice question

y------> the number of free response question

we know that

-----> equation A

-----> equation B

Substitute equation B in equation A

Find the value of x

therefore

the answer is

the number of multiple choice question are

the number of free response question are

Answer:

Given function:

[tex]f(x)=6x^2+x-2[/tex]

To find the zeros of the function, set the function to zero and factor:

[tex]\implies 6x^2+x-2=0[/tex]

[tex]\implies 6x^2+4x-3x-2=0[/tex]

[tex]\implies 2x(3x+2)-1(3x+2)=0[/tex]

[tex]\implies (2x-1)(3x+2)=0[/tex]

Therefore, the zeros are:

[tex]\implies (2x-1)=0 \implies x=\dfrac{1}{2}[/tex]

[tex]\implies (3x+2)=0 \implies x=-\dfrac{2}{3}[/tex]

If α and β are the zeros of the function:

Question 1

[tex]\begin{aligned}\implies \alpha^2+\beta^2 & =\left(\dfrac{1}{2}\right)^2+\left(-\dfrac{2}{3}\right)^2\\\\& = \dfrac{1}{4}+\dfrac{4}{9}\\\\& = \dfrac{9}{36}+\dfrac{16}{36}\\\\& = \dfrac{25}{36}\end{aligned}[/tex]

Question 2

[tex]\begin{aligned}\implies \dfrac{1}{\alpha}+\dfrac{1}{\beta} & = \dfrac{1}{\frac{1}{2}}+\dfrac{1}{-\frac{2}{3}}\\\\& = 1 \times \dfrac{2}{1}+1 \times -\dfrac{3}{2}\\\\& = 2 - \dfrac{3}{2}\\\\& = \dfrac{1}{2}\end{aligned}[/tex]

Answer:

[tex]1) ~\alpha^2+\beta^2 = \dfrac{25}{36}\\\\\\2)~\dfrac 1 {\alpha} + \dfrac 1{\beta} = \dfrac 12[/tex]

Step-by-step explanation:

[tex]\text{Given that,}\\\\f(x) = 6x^2 +x -2~ \text{and the roots are}~ \alpha, \beta\\\\\text{Now,}\\\\\alpha + \beta = -\dfrac{b}{a} = -\dfrac{1}{6}~~~~~;[\text{Compare with the standard form}~ ax^2 +b x + c = 0]\\\\\alpha \beta = \dfrac ca = -\dfrac2 6 = - \dfrac 13\\\\\\\textbf{1)}\\\\\alpha^2 +\beta^2\\\\\\=\left(\alpha +\beta \right)^2 - 2 \alpha \beta \\\\\\=\left( -\dfrac 16 \right)^2 -2 \left(- \dfrac 13 \right)\\\\\\=\dfrac{1}{36}+\dfrac 23\\\\\\=\dfrac{25}{36}[/tex]

[tex]\textbf{2)}\\\\\dfrac 1{\alpha} + \dfrac 1{\beta} \\\\\\=\dfrac{\alpha + \beta}{\alpha \beta }\\\\\\=\dfrac{-\tfrac16}{-\tfrac 13}\\\\\\=\dfrac{1}{6} \times 3\\\\\\=\dfrac{1}{2}[/tex]

The dimensions of the rectangular plot of the land is 17 × 12 square units.

The perimeter of a shape is the distance around the shape. It is the total length of the shape's sides.

Perimeter of rectangle = 2( length + width)

The area can be defined as the space occupied by a flat shape or the surface of an object.

Area of rectangle = length × width

According to the given question.

Let the length and width of the rectangular plot be x and y respectively.

Kadeem has 58m of fencing.

⇒ Perimeter of the rectangular plot of land = 58m

⇒ 2(length + width) = 58

⇒ x + y = 29...(i)

Also, area of plot land = 204 square units.

⇒ x × y = 204

⇒ x × (29 - x) = 204

⇒ [tex]29x -x^{2} = 204[/tex]

⇒ [tex]x^{2} -29x + 204=0[/tex]

⇒ [tex]x^{2} -12x -17x +204=0[/tex]

⇒ [tex]x(x-12)-17(x -12)=0[/tex]

⇒ [tex](x -12)(x -17) = 0[/tex]

⇒ x = 12 or x = 17

When, [tex]x = 12[/tex]

[tex]y = 29-12 = 17m[/tex]

And when, [tex]x = 17[/tex]

[tex]y = 29 - 17 = 12m[/tex]

Hence, the dimensions of the rectangular plot of the land is 17 × 12 square units.

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Answer:

x = 26

Step-by-step explanation:

4(x+7) = 5x+2       1. (4) with x and 7

4x+28 = 5x+2      2. -5x on both sides

-x+28 = 2             3. -28 on both sides

-x = -26                4. /-1 on both sides

x = 26

Answer:

Domain of the function

Step-by-step explanatio

Answer:

i4=1

Step-by-step explanation:

Explanation: Rewrite i14 as (i4)3×i2 . If i=√−1, then i2=−1 . From here (i2)2=(−1)2 , so i4=1

Answer:

a=4m

b=63x¹2 - 35x6 maybe you formatted/typed this wrong, but this is impossible..?

c=4

From the above function, it is clear that the value of f is never 0. Hence the statement that is true is (Option E), See explanation of same below.

Note that the function is related to Euler's number which is depicted as:
e ≈ 2.7182. The function is given as:

f(x) = 100 * [tex]e^{-x}[/tex]

Assuming x = -2, we'd have:

100 * 2.7182[tex]^{-2}[/tex]

= 271.82[tex]^{-2}[/tex]

= 0.00001353354

Hence, even when x tends < 0 the function f(x) thus, is never 0. See the attached graph for confirmation.

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Answer:

[tex]8^{14}[/tex]

Step-by-step explanation:

When we multiply exponents with the same base, the product it the base to the power of the exponents added together. In this case, the base will remain as 8 and the exponent will be 10 + 4 = 14. Therefore, the answer is [tex]8^{14}[/tex].

Answer:

[tex]\large\boxed{\sf \sf 8^{14}}[/tex]

Explanation:

[tex]\rightarrow \sf 8^{10} \ \ x \ \ 8^4[/tex]

apply exponent rule: [tex]\sf \bf a^b\cdot \:a^c=a^{b+c}[/tex]

[tex]\rightarrow \sf 8^{10+4}[/tex]

[tex]\rightarrow \sf 8^{14}[/tex]

Answer:

1. P(2 or 4) - 2/3

2. P(multiple of 2) - 1

3. P(multiple of 3) - 1/3

4. P(odd prime numbers) - 0

Step-by-step explanation:

Hope this helps

P(2 or 4) = 2/3

p(multiple of 2) =  1

p(multiple of 3) = 1/3

P(odd prime numbers) = 0

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

P(2 or 4) = (number of sides that have a value of 2 / total number of sides) + (number of sides that have a value of 2 / total number of sides) = 2/6 + 2/6 = 2/3

p(multiple of 2) = number of sides that are a multiple of 2 / total number of sides

6/6 = 1

p(multiple of 3) = number of sides that are a multiple of 3/ total number of sides

= 2/6 = 1/3

P(odd prime numbers) = number of odd prime numbers / total number of sides  = 0

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Based on the scenario above, the inequalities that can be graphed on the coordinate plane is y < 3x + 5.

In graphing inequalities, one can graph inequality on the coordinate plane.

Note that the solutions for a linear inequality can be seen in the area of the coordinate plane. The boundary line used are  > ,<, ≥ and ≤.

Note also that the  inequalities graphed on the coordinate plane known to be y < 3x + 5 is shown in the image attached. When you look at option A, B and D, you will see that they are incomplete and missing some values but option C gave the correct example of an inequalities.

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Answer:

Hi,

Step-by-step explanation:

I don't know the name in Enghish but in French it is "une règle de 3"

(may be linear rule)

arc length= 2*π*r/360*100=50*π/9

area= π*r²*100/360=250*π/9

Answer:

  dilation by a factor of 1/3 about the origin

Step-by-step explanation:

Each side of triangle B is 1/3 the length of the corresponding side of triangle A. This tells you dilation using a scale factor of 1/3 is involved.

Lines joining corresponding points of triangles A and B converge on the origin. This tells you the center of dilation is the origin.

The one transformation that maps A to B is ...

  dilation by a factor of 1/3 about the origin

_____

Additional comment

No rotation, reflection, or translation are needed, since the triangles have the same orientation, and dilation fully accounts for their relative position.

=======================================================

Explanation:

The range is {2, -1, 5, -8, 9} which sorts to {-8, -1, 2, 5, 9}. This is the set of y coordinates of the points given. Each point is of the form (x,y).

Since the domain and range are the same set in this case, we must have those value mentioned as x coordinates of the points.

The domain is the set of x coordinates of the points given.

The values -8, -1, 2, and 9 are already used as x coordinates. The only thing missing is the 5.

Answer:

30

Step-by-step explanation:

Answer:

51

Step-by-step explanation:

when n=1

when n=2

t2=4

as first term is 1 and second term is 4 so the common difference is 3 ,for 6 term if you put this in formula for sum to the n term you will get 51

t1 =1

d=3

n=6

than S6=51

Answer: 3

Step-by-step explanation:

Opposite sides of a parallelogram are congruent, so AD=BC.

[tex]3x+5=8x-10\\5=5x-1015=5x\\x=\boxed{3}[/tex]

The calculation of the mean shows that Jamie should not lose her job as she was faster after the talk from the boss.

The mean of the travel time in the first scenario will be:

= (18, 20, 22, 27, 16, 23, 25,26, 19, 28)/10

= 224/10

= 22.4

She she shows consistency, the mean will be:

= (22, 20, 22, 24, 24, 23, 25, 21, 19, 21)/10

= 221/10

= 22.1

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Answer:

Step-by-step explanation:

The order of the transformations affects the values used for translation. Here, the translation is described before the reflection and compression.

__

The grayed-out graph of the square root function opens to the right. The darker graph of the transformed function opens to the left, so the reflection is across the y-axis.

__

The square root function goes through the point (1, 1), that is, 1 unit right of the vertex, and 1 unit up. The transformed function goes through the point (1, -1/2), that is, 1 unit left of the vertex and 1/2 unit up. The vertical height of 1 unit on the original graph is compressed to a height of 1/2 unit on the transformed graph. Vertical compression is by a factor of 1/2.

__

The description of the transformation gives the translation before the reflection and compression. So, to find where the uncompressed and unreflected vertex is, we need to reverse those transformations.

Expanding the transformed square root graph by a factor of 2 will undo its compression by a factor of 1/2. That will move the vertex from (2, -1) to (2, 2×(-1)) = (2, -2).

Reflecting this expanded function back across the y-axis will undo the original reflection. That moves the vertex from (2, -2) to (-2, -2). This is where the original translation left the function's vertex before the reflection and compression moved it to the location shown.

The translation is right -2 and down 2.

_____

Additional comment

It is possible that you will get pushback on these answers. If so, you should have your teacher demonstrate the transformations in the order described.

__

In the order described, we have ...

  [tex]f(x)=\sqrt{x}\\\\f_1(x)=\sqrt{x-(-2)}=\sqrt{x+2}\qquad\text{translation right -2}\\\\f_2(x)=\sqrt{x+2}-2\qquad\text{translation down 2}\\\\f_3(x)=\sqrt{-x+2}-2\qquad\text{reflection in the y-axis}\\\\g(x)=\dfrac{1}{2}(\sqrt{-x+2}-2)\qquad\text{compression vertically by $\dfrac{1}{2}$}[/tex]

The attached graph shows the result of this sequence of transformations.

__

If the reflection and compression were done before the translation, then the translation would be 2 right and 1 down.

Answer:

D. g(x) = -x² - 2

Explanation:

Parent function: x²

Graph g(x) is inverted so -x² and then shifted down 2 units so -x² - 2

Hence: g(x) = -x² - 2

Answer:

5x²+12x+4

Step-by-step explanation:

5x²+10x+2x+4

5x²+12x+4

Answer:

[tex]5x^{2}+12x+4[/tex]

Step-by-step explanation:

Use the FOIL method to expand.

5x*x+

2*x+

5x*2+

2*2

which is

5x^2+2x+10x+2

combining the two x s we get

5x^2+12x+4

Answer:

Step-by-step explanation: If I have to Write In Slope- Intercept form, This Is how I explain it.

Since x = −3 is a vertical line, there is no y-intercept and the slope is undefined.

Slope: Undefined

y-intercept: No y-intercept

Answer:

Step-by-step explanation:

According to the diagram we observe:

As we know the exterior angle is same as the some of the remote interior angles.

It can be shown as:

Substitute the values and solve for x:

Find the measure of m∠ABC:

The matching answer choice is D.