A ride-share company has a fee that includes a fixed cost and a cost that depends on both the time spent travelling, in minutes, and the distance travelled, in kilometres. The fixed cost of a ride is $2.55 Judy's ride costs $16.75 and took eight minutes. The distance travelled was 10 km. Pat's ride cost $30.35 and took 20 minutes. The distance travelled was 18 km. Roy's ride took 10 minutes. The distance travelled was 15 km. Find the cost of Roy's ride.

Answer:

CP = ₹ 700

Step-by-step explanation:

Let the cost price be 'x'

SP = [tex](\frac{100+Profit}{100})*CP[/tex]

   [tex]= \frac{100+10}{100}*x\\\\=\frac{110}{100}*x\\\\= \frac{11}{10}x[/tex]

If it had been sold are 10% loss, then

SP = [tex](\frac{100-loss}{100})*CP[/tex]

     [tex]=\frac{100-10}{100}*x\\\\=\frac{90}{100}*x\\\\=\frac{9}{10}x[/tex]

Difference in both the selling price =  140

[tex]\frac{11}{10}x-\frac{9}{10}x=140\\\\\\\frac{2}{10}x=140\\\\\frac{1}{5}x=140\\\\ x = 140*5\\x = 700[/tex]

Answer:

132.8 are left enjoy

Answer:

The two column proof is presented as follows;

Statement  [tex]{}[/tex]                            Reason

ΔSCW ≅ ΔTUW [tex]{}[/tex]                    Given

[tex]\overline {SV}[/tex] ≅ [tex]\overline {TU}[/tex] [tex]{}[/tex]                               CPCTC

[tex]\overline {SW}[/tex] ≅ [tex]\overline {TW}[/tex]  [tex]{}[/tex]                            CPCTC

[tex]\overline {WV}[/tex] ≅ [tex]\overline {UW}[/tex] [tex]{}[/tex]                            CPCTC

∠SVW ≅ ∠TUW  [tex]{}[/tex]                   CPCTC

SU = SW + UV [tex]{}[/tex]                       Additive property of Length

TU = TW + VW [tex]{}[/tex]                       Additive property of Length

SU = TW + VW [tex]{}[/tex]                       Substitution

SU = TV [tex]{}[/tex]                        [tex]{}[/tex]          Transitive property

ΔSTV ≅ ΔTSU  [tex]{}[/tex]                       SAS

∠TSV ≅ ∠STU [tex]{}[/tex]                        CPCTC

Step-by-step explanation:

The two column proof is presented as follows;

Statement  [tex]{}[/tex]                            Reason

ΔSCW ≅ ΔTUW [tex]{}[/tex]  Given

[tex]\overline {SV}[/tex] ≅ [tex]\overline {TU}[/tex] [tex]{}[/tex]            Congruent Parts of Congruent Triangles are Congruent

[tex]\overline {SW}[/tex] ≅ [tex]\overline {TW}[/tex]  [tex]{}[/tex]         Congruent Parts of Congruent Triangles are Congruent                  

[tex]\overline {UV}[/tex] ≅ [tex]\overline {VW}[/tex] [tex]{}[/tex]         Congruent Parts of Congruent Triangles are Congruent                  

∠SVW ≅ ∠TUW  [tex]{}[/tex]Congruent Parts of Congruent Triangles are Congruent                  

SU = SW + UV [tex]{}[/tex]    Additive property of Length

TU = TW + VW [tex]{}[/tex]   Additive property of Length

SU = TW + VW [tex]{}[/tex]    Substitution

SU = TV [tex]{}[/tex]              Transitive property

ΔSTV ≅ ΔTSU  [tex]{}[/tex]   Side-Angle-Side rule of congruency

∠TSV ≅ ∠STU [tex]{}[/tex]     Congruent Parts of Congruent Triangles are Congruent                  

Answer:

q ≥ 5/7

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

-6q + 7 ≤ 8q - 3

Step 2: Solve for q

Here we see that any number q greater than or equal to 5/7 would work as a solution to the inequality.

Answer:

q ≥ [tex]\frac{5}{7}[/tex]

Step-by-step explanation:

Given

- 6q + 7 ≤ 8q - 3 ( add 6q to both sides )

7 ≤ 14q - 3 ( add 3 to both sides )

10 ≤ 14q ( divide both sides by 2 )

5 ≤ 7q ( divide both sides by 7 )

[tex]\frac{5}{7}[/tex] ≤ q , then

q ≥ [tex]\frac{5}{7}[/tex]

Answer:

26/63

Step-by-step explanation:

Try to make the denominators equal so that you can add them:

4(3)/21(3)= 12/63

2(7)/9(7)= 14/63

And now you add, which will give you:

12/63 + 14/63= 26/63

Answer:

[tex]A^{I}[/tex]= (7, -1)

Step-by-step explanation:

Rotation is a method of rigid transformation in which a point is moved about a reference point through a stated angle. The rotation could be clockwise or counterclockwise so as to produce the image of the initial object.

In the polar coordinate given, the coordinate of A is given as;

A = (3, 3)

On rotating point A about point C counterclockwise for [tex]270^{o}[/tex], an image [tex]A^{I}[/tex] would be produced.  

But [tex]A^{I}[/tex] would have a coordinate;

[tex]A^{I}[/tex]= (7, -1)

The image of A produced is [tex]A^{I}[/tex] (7, -1).

Answer:

Average height = 64.2 inches

Step-by-step explanation:

[tex]Average = \frac{sum \ of \ the \ values }{number \ of\ value}[/tex]

            [tex]=\frac{62 + 70 + 60 + 63 + 66}{5}\\\\=\frac{321}{5}\\\\=64.2 \ inches[/tex]

The average of the given heights is 64.2 inches

An average of a list of data is the expression of the central value of a set of data. It is defined as the ratio of summation of all the data to the number of units present in the list.

According to the given problem,

Given data = { 62, 70, 60, 63, 66 }

Sum of the given heights = ( 62 + 70 + 60 + 63 + 66 )

                                          = 321

Average = [tex]\frac{321}{5}[/tex]

              = 64.2 inches

Hence, we can conclude, the average of the given data is 64.2 inches.

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Answer:

A=x-3

B=X-14

C=X-10

Answer:

12m²

Step-by-step explanation:

are the files to be used only for the floor, or also for the walls ? ceiling ?

if it is only for the floor, then the height of the room is irrelevant.

perpendicular clearly stands here for rectangular.

so, the good of the room is a rectangle of 4m length and 3m width.

the area of a rectangle is length×width.

area = 4×3 = 12m²

so, 12m² of tiles will be needed (in reality there would be at least 10% to 20% added for bad tiles, wrong cuts, ...).

if walls need to be included, then you will have to add the areas of 4 more rectangles of length×height.

which would be 2 times 3×2.5 = 7.5m² and 2 times 4×2.5 = 10m².

so, the total area incl. walls would be

12 + 7.5 + 7.5 + 10 + 10 = 47m²

if the ceiling has to be included too, then you need to add the area of another rectangle of the same size as the floor.

so, that total would then be

47 + 12 = 59m².

Answer:

second option is correct

Answer:

Area = 160 cm²

Perimeter = 55.31 cm

Step-by-step explanation:

Applying,

Area of the figure = area of the rectangle- area of the triangle

A = lw+bh/2............... Equation 1

From the question,

Given: l = 16 cm, w = 12 cm, h = 8cm, b = 8 cm

Substitute these values into equation 1

A = (16×12)-(8×8/2)

A = 192-32

A = 160 cm²

Perimeter P = /BC/+/CD/+/DE/+/EF/+/FG/+/GA/

Applying pythagoras theorem to get /FG/

FG² = GE²+EF²

FG² = 8²+8²

FG² = 64+64

FG² = 128

FG = √128

FG = 11.31 cm

Therefore,

P = 16+12+4+8+11.31+4

P = 55.31 cm

Answer:

l

Step-by-step explanation:

6.2a = 0.0143

=> a = 0.0143/6.2

=> a = 0.0023......

Answer:

A

Step-by-step explanation:

The line is parallel to y axis, so the equation is x = 5

Answer:

part 1: $532.95. Part 2: 16

Step-by-step explanation:

Part 1: 15 * 15.35 = 230.25

30 * 10.09 = 302.70

230.25 + 302.70 = $532.95

Part 2: R = letters in the word desk

there are 4 letters in the word Desk

[tex]2^{R}[/tex] = [tex]2^{4}[/tex] = 2 * 2 * 2 * 2 = 4 * 4 = 16 subsets

Answer:

b

Step-by-step explanation:

Answer: B <S

Step-by-step explanation:

Answer:

Domain:  all real numbers

Step-by-step explanation:

The domain is the values that x can take

X can be any real number

Domain:  all real numbers

Given:

Two points are A(6,1) and B(9,4).

To find:

The equation of the line that passes through the given points.

Solution:

If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the equation of the line is:

[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

It is given that the line passes through the two points A(6,1) and B(9,4). So, the equation of the line is:

[tex]y-1=\dfrac{4-1}{9-6}(x-6)[/tex]

[tex]y-1=\dfrac{3}{3}(x-6)[/tex]

[tex]y-1=1(x-6)[/tex]

[tex]y-1=x-6[/tex]

Adding 1 on both sides, we get

[tex]y-1+1=x-6+1[/tex]

[tex]y=x-5[/tex]

Therefore, the equation of the line is [tex]y=x-5[/tex].

Answer:

24.6

Step-by-step explanation:

take 72 degree as reference angle

using tan rule

tan 72 = opposite / adjacent

3.07 = x/8

3.07 * 8 = x

24.56 = x

24.6 = x

Answer:

Step-by-step explanation:

[tex]\frac{a^{m}}{a^{n}}=a^{m-n}\\\\\frac{3n^{5}}{6n^{3}}=\frac{3}{6}*n^{5-3}\\\\=\frac{1}{2}n^{2}[/tex]

Answer:

[tex]\frac{n^{2} }{2}[/tex]

Step-by-step explanation:

Rewrite the division as a fraction.

[tex]\frac{3n^{5} }{6n^{3} }[/tex]

Cancel the common factor of 3 and 6.

[tex]\frac{3(n^{5}) }{6n^{3} }[/tex]

[tex]\frac{n^{5} }{2n^{3} }[/tex]

Cancel the common factor

[tex]\frac{n^{2} }{2}[/tex]

Hope this helps!

If you need more help with questions like these please reach out to me!

Answer:

Question 1: 75 minutes

Question 2: 4:07

Question 3: 30 minutes

Question 4: 8:35

Step-by-step explanation:

An hour is sixty minutes; that's the key to solving these questions.

The time basketball was played is 75 minutes or 1 hour 15 minutes

The time I get home is at 4:07

The cook was cooked for 30 minutes.

The cheerleading was over at 8:35.

It is the conversion of one unit to another unit with its standard conversion.

We have,

Trey and Na Asia played basketball from 10:30 - 11:45.

From 10 to 11 we have 60 minutes.

= 60 minutes - 30 minutes [10:30] + 45 minutes [ 11:45]

= 105 minutes - 30 minutes

= 75 minutes

The time basketball was played.

= 75 minutes

= 1 hour 15 minutes

I left school at 3:15 on the bus. I got home 52 minutes later.

1 hour = 52 minutes

= 3:15 + 52 minutes

= 3  + 15 minutes + 52 minutes

= 3 + 67 minutes

= 3 + 1 hour +  7 minutes

= 4:07

The time I get home is at 4:07

Antavious makes Macaroni and cheese. He begins making it at 6:35 and finished at 7:05.

= 6:35 to 7:05

From 6 to 7 we have 60 minutes.

So,

= 60 minutes - 35 minutes [6:35] + 5 minutes [7:05]

=  65 minutes - 35 minutes

= 30 minutes

The cook was cooked for 30 minutes.

Makhiya and Madison went to cheerleading at 7:15. It was over 1 hour and 20 minutes later.

= 7:15 + 1 hour + 20 minutes

= 7 + 15 minutes + 1 hour + 20 minutes

= 8 + 35 minutes

= 8:35

The cheerleading was over at 8:35.

Thus,

The time basketball was played is 75 minutes or 1 hour 15 minutes

The time I get home is at 4:07

The cook was cooked for 30 minutes.

The cheerleading was over at 8:35.

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Answer:

∠B = 74°

a ≈ 14.5 inches

c ≈ 11.4 inches

Step-by-step explanation:

The given parameters of triangle ΔABC are;

∠A = 62.2°, ∠C = 43.8° and side [tex]\overline{AB}[/tex] = b = 15.8 in

By sine rule, we have;

[tex]\dfrac{a}{sin(\angle A)} = \dfrac{b}{sin(\angle B)} = \dfrac{c}{sin(\angle C)}[/tex]

[tex]a = {sin(\angle A)} \times \dfrac{b}{sin(\angle B)}[/tex]

By angle sum property, we have;

∠B = 180° - (∠A + ∠C)

∴ ∠B = 180° - (62.2° + 43.8°) = 74°

∠B = 74°

[tex]\therefore {a} = {sin(62.2^{\circ})} \times \dfrac{15.8}{sin(74^{\circ})} \approx 14.5[/tex]

a ≈ 14.5 in.

[tex]c = {sin(\angle C)} \times \dfrac{b}{sin(\angle B)}[/tex]

[tex]\therefore {c} = {sin(43.8^{\circ})} \times \dfrac{15.8}{sin(74^{\circ})} \approx 11.4[/tex]

c ≈ 11.4 in.

The mode is the highest frequency data among the given data set is 2.

The value that appears most frequently in a data set is called the mode. One mode, several modes, or none at all may be present in a set of data.

Mode is the highest frequency of any number among data

For example 2,3,5,5,6,5,9,8,5,1

Here mode will be 5 because it has a frequency of 4 remaining all have a lower frequency than 5.

Given the data set is

2, 8, 16, 4, 12, 9, 2

2 has frequency = 2

And remaining all have a frequency as 1

Since 2 has the highest frequency so it will be mode.

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Answer:

2103 bags

Step-by-step explanation:

8412/4=2103

Answer:

pretty sure its 40cm

Step-by-step explanation:

hope this helps :)

Answer:

36cm

Step-by-step explanation:

Tan(14/12)=angle of ONP

Tan(42/X)=angle of LNM

singe those angels are the same;

Tan(14/12)=Tan(42/X)

14/12=42/X

14*X=42*12

X=(42*12)/14

X=36

Answer:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!;;;;!!!!!!!!!!!

The required equation of graph f(x) is y = -x²-3.

A parabola is a U-shaped plane curve where any point is at an equal distance from a fixed point (known as the focus) and from a fixed straight line which is known as the directrix.

here, we have,

Given that,

The graph of f(x), resembles the graph of G(x) =x^2 .

We have to find,

Which of the following could be the equation of  f(x).

According to the question,

Function, G(X) = x^2 ,

The graph in the given question figure. It is a parabola with vertex at (0,0).

By the graph predict that Sign of  x^2 is positive, that is why the parabola opens up.

General equation of parabola is given by,

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

Where, a = 1, and vertex (h, k) is (0,0).

The graph of F(x) unknown function opens down ,

Sign of x^2 as negative. and the vertex is at (0,-3)

To find the equation of function f(x),

Putting the values of a and vertex coordinates,

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

The equation of parabola become:

y = -x²-3.

(solving we get)

Hence, The required equation of f(x) is y = -x²-3.

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Answer:

A. 120

Step-by-step explanation:

600÷60=10

10×12=120

Hope this helps! :)