Consider the line y=9x-1.
Find the equation of the line that is perpendicular to this line and passes through the polnt (6, -5).
Find the equation of the line that is parallel to this line and passes through the point (6,-5).

The triangles ABC and EDC have corresponding sides

The triangles ABC and EDC are similar

The given parameters are:

ED = 8, DC = 10, EC = 12, AB = 12

BC = 15 and AC = 21

To determine that the triangles are similar, we simply divide the corresponding sides to determine the scale factor (k).

So, have:

k = AB/ED = BC/DC

This gives

k = 12/8  = 15/10

Evaluate the quotients

k = 1.5 = 1.5

The scale factors are equal

Hence, the triangles ABC and EDC are similar

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The total number of people attending the game is 1700.

Percentage in simple terms can be defined as a part per 100.

Percentage is a ratio or number which has to be defined as a fraction of 100.It is denoted by %.

If we are asked to calculate the percent of a number simply divide it by the whole part or number and then multiply with 100.

Percent of people supporting the home team = 92%

Number of people supporting the home team = 1564

Let x be the total number of people attending the game.

⇒ [tex]\frac{1564}{x}[/tex] × 100 = 92

⇒ [tex]\frac{1564}{x}[/tex] = 0.92

⇒ x = [tex]\frac{1564}{0.92}[/tex]

⇒ x = 1700

Hence, the total number of people attending the game will be 1700.

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Answer:

yes

Step-by-step explanation:

your answer is correct

Answer:

115

Step-by-step explanation:

If 65 is subtracted from 180the answer is 115

Answer:

The correct polynomial is F. Calculation errors combining h terms and distributing -1.

Step-by-step explanation:

[tex]( - 4h + 3)( {h}^{2} + 9h - 1)[/tex]

[tex] - 4 {h}^{3} - 36 {h}^{2} + 4h + 3 {h}^{2} + 27h - 3[/tex]

[tex] - 4 {h}^{3} - 33 {h}^{2} + 31h - 3[/tex]

[tex]\qquad \qquad \huge \pink {\sf{☁Answer☁}} \\ \\ [/tex]

[tex] \large \purple { \rm{Let's \: solve→}} \\ [/tex]

[tex] \rm{3(2x-5)=4x+33} \\ \\ \rm {solving \: bracket} \\ \\ \rm{6x - 15= 4x + 33} \\ \\ \rm{arranging \: terms} \\ \\ \rm{6x - 4x = 33 + 15} \\ \\ \rm{2x = 48} \\ \\ \rm{x = \frac{48}{2} } \\ \\ \rm{x = \cancel \frac{48}{2} } \\ \\ \rm{x = 24}[/tex]

[tex]\purple{\rule{15mm}{2.9pt}} \red{\rule18mm{2.5pt}} \orange{ \rule18mm{2.5pt}}[/tex]

[tex]\sf{\:мѕнαcкεя\: ♪...}[/tex]

Answer:

[tex]\fbox{X = 24}[/tex]

Solution:

Given expression ➟ 3(2x-5) = 4x + 33,

using distributive property open the above bracket,

➳ 3×2x - 3×5 = 4x + 33

Using BODMAS solve the L.H.S,

➳ 6x - 15 = 4x + 33

now arrange the like terms aside,

➳ 6x - 4x = 33 + 15

➳ 2x = 48

➳ x = 48/2

➳ x = 24

the given expression is true for x = 24

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Answer:

When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is transformed into its opposite (its sign is changed).

Step-by-step explanation:

hope this helps if does mark brainllest

Answer: £198.27

Step-by-step explanation:

Answer:

Step-by-step explanation:

In general, a scale factor is applied to an original (pre) to obtain an image (post). It is the ratio post/pre that expresses the relationship between image dimensions and original dimensions.

Here, that post/pre ratio is 3/4 = 0.75.

So

f(2)

f(3)

f(4)

Answer:

square root of 39

Step-by-step explanation:

5^2+b^2=8^2

25+b^2=64

64-25=39

Check the picture below.

let's notice something, the equation of line B is in slope-intercept form, thus

[tex]y = 2x + \underset{\stackrel{\uparrow }{b}}{2}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]

so using that grid scale, we can just get two points from the line A to get its equation, let's use those in the picture below

[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-6)}}}\implies \cfrac{0+6}{6+6}\implies \cfrac{1}{2}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y+6=\cfrac{1}{2}(x+6)\implies y+6=\cfrac{1}{2}x+3\implies y=\cfrac{1}{2}x-3[/tex]

Answer:Eight to 12 inches is usually adequate. If drainage is a problem, or if the plants you are growing prefer drier soil, the bed could be taller and filled with a porous growing medium.

Step-by-step explanation:

Answer:

This is hard

Step-by-step explanation:

I'm so sorry I can't do it

Answer:

[tex]\frac{69}{20}[/tex]

Step-by-step explanation:

[tex]\frac{3}{7} -\frac{11}{4} +\frac{16}{5} +\frac{18}{7}[/tex]

= [tex]\frac{3}{7} +\frac{18}{7}-\frac{11}{4} +\frac{16}{5}[/tex]

= [tex]\frac{21}{7} -\frac{11}{4} +\frac{16}{5}[/tex] As both denominators have 7 you can add their numerators

=[tex]\frac{21}{7} -(-\frac{9}{20})[/tex]     L.C.M

=[tex]\frac{21}{7} +\frac{9}{20}[/tex]   L.C.M

=[tex]\frac{69}{20}[/tex]

Answer:

3*30 = 90

Step-by-step explanation:

3*30 = 90

Answer:

[tex]b^2 = 14[/tex]

Step-by-step explanation:

log(base)(power) = (exponent).

In this case: logb(14) = 2

Note the variable/number in place of each term. Set the equation:

base^power = exponent

[tex]log_b 14 = 2\\b^2 = 14[/tex]

[tex]b^2 = 14[/tex] is your answer.

~

Answer:

See below

Step-by-step explanation:

.2 of an hour   is     .2 * 60 min = 12 min

120 min / 10 times = 12 min per trip

The train takes 12 minutes to go around the museum once.

A mathematical arithmetic operation is multiplication. Moreover, it is the practice of repeatedly adding the same expression kinds.

Example: 2 x 3 means that 2 is multiplied by 3 or that 3 is multiplied by 2 times.

A toy train that travels the length of a train museum is present. In two hours, the train travels roughly ten times.

The error in the student's work is converting 0.2 hours to 20 minutes. 0.2 hours is equal to 12 minutes, not 20 minutes.

To correct the student's work, we can multiply 0.2 by 60 to convert it to minutes:

0.2 hours × 60 minutes/hour = 12 minutes

Therefore, it takes 12 minutes for the train to go around the museum 1 time.

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Answer: 21/572

[tex]\frac{5.5}{10 . 22} - \frac{4}{52}[/tex]

The dots stand for multiplication

Answer:

5. radius= TD

inscribed angle =RQS

diameter =TA

6. radius=SY

secant=TU

Tangent=S

Answer:

22 is th area

Step-by-step explanation:

Answer:

Area of cube = 6l²

= 6×2²

= 24 ft²

2 feet.

how many square feet of wood he used to build the box.

each side measures 2ft.

A=L²=(2 ft)²=4 ft²

A cube has 6 sides so the total quantity of wood is:

Wood=6×A

=6×4 ft²

=24 ft²

he used 24 ft² of wood to build the box.

Step-by-step explanation:

to figure this out you would have to take all the numbers and do the rise over run technique

Answer:

c and d

Step-by-step explanation:

because

- In f(x),evey time you find  X you write(-6x+5) instead

4x²-----------4(-6x+5)²

3x+1------------3(-6x+5) +1

-In g(x),evey time you find  X you write(-4x² +3x+1)  instead          

The ways that the policies of President Kennedy and President Johnson are similar and different is given below:

After the assassination of Kennedy, Johnson was known to have followed the footsteps of his predecessor's and went ahead on implementing some of his programs.  Johnson was known to have went ahead with the the anti-poverty program and the Civil Rights Bill.

The differences between the both presidents were:

Conclusively, these two president were said to be very were similar, and yet they were different.

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Answer:

1. B = 6 m²

2. h = 2 m

3. V = 12 m³

Step-by-step explanation:

the base is a right triangle

Then

[tex]B=\frac{3\times 4}{2} =6\ m^{2}[/tex]

____________

The height of the prism is the length of the segment that join the base to its opposite face

Then

h = 2m (given)

___________

The volume of the prism :

[tex]V=B\times h=6\times 2=12 m^3[/tex]

Answer:2.4 isn’t occupied, whilst 13.3 is full

Step-by-step explanation: answerrr

Step-by-step explanation:

It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.

So, let the two positive integers be x and y.

Therefore,

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]

Now, From the first equation we have,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]

Now, substituting the value of y in equation (ii) we get :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]

Now using the quadratic formula :

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]

Where,

[tex] \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]

Now, we have two equations,

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]

So, Equation (iv) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]

Now, Equation (v) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]

Now, we are to find the value of y.

Substituting the value of x (24) in equation (iii) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]

Again, Substituting the value of x (7) in equation (iii) :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]

Therefore,

[tex] \\ [/tex]

So, The smaller of the numbers is 7 .

The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.

This question belongs to quadratic equations so we have to find the answer by making equation and solving it .

According to question , sum of two positive integers is 31 . So ,

According to question , sum of square

of these numbers is 625 . So ,

From equation 1 ( x + y = 31 ) , Value

of x :

Now , putting value of x in eq. 2 :

Solving it by using middle term

splitting :

First number ,

Second Number ,

According to question ,

Sum of numbers is equal to 31 :

Sum of squares of these numbers is equal to 625 :