If a line has a slope of 2 and contains the point (-2, 1), what is its equation in
point-slope form?

A. y +2=-2(x-1)
B. y-2 =-2(x-1)
C. y-1=2(x-2)
D. y- 1 = 2(x+ 2)

There appear to be 25% more sales of horror books than cookbooks, but there are really only about 2% more.

Percentage is a measure of frequency that expresses the ratio of two numbers as a value out of 100.

Percentage of horror books to comedy books = [(number of horror books sold / number of comedy book) - 1 ] x 100

[(55 / 54) - 1] x 100 = 2%

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Answer: 1899.70 cu ft

Step-by-step explanation:

Cone volume = 1/3 TT r^2 h

V= (1/3) (3.14) (11^2) (15)

V=1899.70 cu ft

The addition of the given fractions ( 1/6 + 2/4 ) to its simplest fraction form is 2/3.

Given that;

1/6 + 2/4

First, multiply the denominators then cross-multiply nominators and denominators.

1/6 + 2/4 = ( (6×2) + ( 4×1 ) ) / 6 × 4

1/6 + 2/4 = ( 12 + 4 ) / 24

1/6 + 2/4 = 16 / 24

We simplify

1/6 + 2/4 = 16 / 24 = 2/3

Therefore the addition of the given fractions ( 1/6 + 2/4 ) to its simplest fraction form is 2/3.

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

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:

It is 13 and 8. Look down below and you will know the answer is right:

Step-by-step explanation:

10x+50(21-x)=530

10x+1050-50x=530

-40x=530-1050

-40x=-520

x=-520/-40

x=13

10 cent coin 13

50 cent coin 8

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Answer:

maybe the answer is x-y =-5

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

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

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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Answer:  B. They are one-dimensional objects

This applies to line segments and lines as well.

One dimensional objects are such that they only have one component to them when it comes to determining a location on that object.

For instance, a number line is one dimensional because we only need one number to determine where a location is.

A ray is effectively a half-line, or put another way, a line is the combination of two rays glued together. Basically a ray points forever in one direction only, as a line will point in two opposite directions.

Extra side notes:

Answer: B

Step-by-step explanation:

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: [tex]12\sqrt{5}[/tex]

Required theorem:

Pythagorean theorem = [tex]a^2 + b^2=c^2[/tex]

In other words, one side length squared + another side length squared = the middle lined squared.

Step-by-step explanation:

[tex]60^2+60^2 = c^2\\\\360+360=c^2\\\\720=c^2\\\\\ \sqrt{c^2} =\sqrt{720}\\\\c=+/- 12\sqrt{5}[/tex]

We can only have a positive distance so your final answer is :

[tex]12\sqrt{5}[/tex]

Answer:

  A. one solution; lines intersect at one point

  B. (x, y) = (3, 4)

Step-by-step explanation:

A linear equation will be represented on a graph by a straight line. A pair of equations will graph as a pair of lines. If the lines are not coincident or parallel, they will intersect at exactly one point. That point is the solution to the pair of equations.

__

A graph of the given points quickly shows the lines are not parallel or coincident. Each point on line A satisfies the equation that describes the line. Each point on line B satisfies the equation that describes line B. The point of intersection is a point that satisfies both equations.

A pair of distinct non-parallel lines can have at most one point of intersection. Hence there can only be one solution to the pair of equations.

__

The graph shows the solution to the equations for lines A and B is ...

  (x, y) = (3, 4)

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:

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.

Answer:

  see attached

Step-by-step explanation:

The two different triangles can be formed by placing the 90° angle adjacent to, or opposite the given side.

__

In the attached diagram, the two triangles are ABC and ABD. The right angles are at vertex C and vertex B, respectively.

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.

[tex]\textbf{Given that,}\\\\~~~\dfrac 9{12} \div \dfrac{2}{12}\\\\\textbf{We can rewrite it as}\\\\~~~~\dfrac{9}{12} \times \dfrac 1{\tfrac{2}{12}}\\\\\\=\dfrac {9}{12} \times \dfrac{12}2\\\\\\=\dfrac 92\\\\\\=4.5\\\\\textbf{Hence the quotient is}~ \dfrac 92 ~ \textbf{or}~ 4.5.[/tex]

The scatterplot shows no correlation, a positive correlation, a negative correlation. Because the values of y  stay the same, increase,  decrease,  as the values of x increase.

Scatterplot shows relation between two quantitative variables measured for the same element.

The negative correlation shows that  y decreases when x increases.

The positive correlation shows that y increases when x decreases.

No correlation shows shows that there is no change in the value of y when x stay same.

Thus, The scatterplot shows no correlation, a positive correlation, a negative correlation. Because the values of y stay the same, increase,  decrease,  as the values of x increase.

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Answer:

Yes, they are congruent. They must be congruent because you know that 3 out of the 3 angles are identical between the two triangles. You know this because both triangles have a right angle (90°) and another identical angle (the two curves). This means that the 3rd angle for both triangles must be identical as well. Therefore, the triangles are the same.

Step-by-step explanation:

plant cells have a cell wall for to prepared their own food but animal cell haven't cell wall

plant cell have a large vacuoles but in plant cell it has Avery small vacuoles the two have the main difference between animal cell plant cell

Answer:

Option D

Step-by-step explanation:

[tex]\frac{2}{3}x=\frac{12}{5}[/tex]

[tex]x=(\frac{12}{5} )(\frac{3}{2} )=\frac{36}{10} =\frac{18}{5} =3\frac{3}{5}[/tex]

Hope this helps

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

Step-by-step explanation:

Do you have a final product for the equation.

I am unable to find the missing part of the peoblome without a total to sum up to.

Sorry...

Answer:

a=4m

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

c=4

[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]