jada has $20 to spend on games and rides at a carnival. Games cost $1 each and rides are $2 each.

which equation represents the relationship between the number of games, x , and the number of rides, y,
that Jada could do if she spends all her money?
A: x + y =20
B: 22 + 4 = 20
C: 0+2y=20

Answer:

Step-by-step explanation:

According to the matrix, the system is:

Correct match is A

Answer:

Option 4: 0

Step-by-step explanation:

Given the function, f(x) = 3x, and the domain. {-3, 1, 4}:

Substitute the values of each domain as inputs into the function:

f(-3) = 3(-3)

f(-3) = -9

f( 1 ) = 3( 1 )

f( 1 ) = 3

f(4) = 3(4)

f(4) = 12

Therefore, the correct answer is option 4) 0, because there is no input value that provides an output of 0. You were only given three domain values that provided its corresponding outputs.

The equivalent expression of the different of two square is -1

Given the following expressions

a cot α = 1

a = 1/cot α = sinα/cosα

b = 1/cos α

We are to find the expression a² - b²

According to difference of two squares;

a² - b² =  (a + b) (a - b)

Substitute the given expressions into the formula as shown:

[tex]a^2 - b^2 = (\frac{sin \alpha}{cos \alpha} )^2 - (\frac{1}{cos \alpha} )^2\\a^2 - b^2=\frac{sin^2 \alpha}{cos^2 \alpha} -\frac{1}{cos ^2 \alpha}\\a^2 - b^2=\frac{sin^2 \alpha-1}{cos^2\alpha}\\a^2 - b^2=\frac{-(1-sin^2\alpha)}{cos^2\alpha} \\a^2 - b^2=\frac{-os^2 \alpha}{cos^2 \alpha}\\a^2 - b^2 = -1[/tex]

This shows that the equivalent expression of the difference of two squares is -1

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Answer:

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Answer:

If the two random variables are independent, the covariance will be zero. It means they don't have any linear relationship.

Step-by-step explanation:

Answer:

There were 200 Questions on the Final.

Step-by-step explanation:

The expression which relates the total time taken and the time taken for the trip is [15s + 15(s+2)] / s² + 2s and 2.32 hours respectively.

Travel time = distance ÷ speed

Second half of the trip :

Time taken for second half of trip :

Time taken = 15 / s

First half of the trip :

Time taken for first half of trip :

Time taken = 15 / (s+2)

Total time taken :

First half + second half

15/(s+2) + 15/s = [15s + 15(s+2)] / s² + 2s

B)

If s = 12

Substitute s = 12 into the expression :

[15(12) + 15(12+2)] / 12² + 2(12)

[180 + 210] / 144 + 24

390 / 168

= 2.32 hours

Therefore, the total time taken is 2.32 hours.

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Answer:

...

Step-by-step explanation:

Answer:

Step-by-step explanation:

A) ∠1 =  ∠2   {Alternate angles are equal}

6x + y - 4 = x - 9y  + 1

6x + y      = x - 9y + 1 + 4

6x + y      = x - 9y + 5

6x - x  + y + 9y = 5

  5x  + 10y = 5

Divide the entire equation by 5

  x  + 2y = 1 -------------(I)

∠2 + ∠3 = 180       {Linear pair}

x - 9y + 1 + 11x + 2 = 180

x + 11x - 9y   = 180 - 2 - 1

          12x - 9y = 177

Divide the whole equation by 3

       4x - 3y  = 59    -------------(II)

B) Multiply equation (I) by 3 and multiply equation (II) by 2. Thus y will be eliminated and we can find the value of x.

(I)* 3           3x + 6y = 3

(II)*2           8x - 6y =  118    {Now add}

                  11x        = 121

                            x = 121/11

x = 11

Plugin x = 11 in equation (I)

                            11 + 2y = 1

                                  2y = 1 -11

                                  2y = -10

                                   y = -10/2

y = -5

C) ∠1 = 6x + y - 4

         = 6*11  - 5 - 4

          = 66 - 5 - 4

           = 57

∠2  = x - 9y + 1

        = 11 -9*(-5) + 1

         = 11 + 45 + 1

          = 57

∠3 = 11x + 2

      = 11*11 + 2

      = 121 + 2

      = 123

Step-by-step explanation:

Total amount is:

If the numbers are correct ₱500 won't be enough as the total amount exceeds it.

Mother bought 3 cans of sardines at ₱17.80 each, 5 cans of milk at ₱38.85 each, 3kg sugar at ₱60.50 each, and 2 cans of corn beef at ₱78.85 each. Find the total amount bought. If she paid ₱500 bill, find her change.

Mother bought 3 cans of sardines at ₱17.80 each, 5 cans of milk at ₱38.85 each, 3kg sugar at ₱60.50 each, and 2 cans of corn beef at ₱78.85 each.

The total amount for which she had brought the items

= ₱ (3×17.80 + 5×38.85 + 3×60.50 + 2×78.85)

= ₱ (53.40 + 194.25 + 181.50 + 157.70)

= ₱ 586.85

The total amount for which she had brought the items is ₱ 586.85.

I cannot solve the second part of the sum because the amount exceeds ₱ 500.

Hope it helps

Answer:

A

Step-by-step explanation:

With way one, we can easily count the tally marks, but with way 2 it becomes much harder for us to count the tallies when we are done.

Answer:

a

Step-by-step explanation:

its easy to count way 1 but way 2 is harder to count

Answer:

The answer is 15m

Step-by-step explanation:

The meaning whole number means an actual whole number, decimals are just part's of a whole number, so excluding decimals:

If you look at 5 and compare it to 3 following next, 3 is not higher then 5 and therefore is rounded down.

Hope this helps!

Answer:

answer is choice 4 because in te given angle A and angle B are congrunt and if they are acute the degree measure is 0-90

Answer:

1) 1085 ÷ 7

7 )1085(155

- 7

38

- 35

× 35

- 35

0

1085 ÷ 7 = 155

2) 7104 ÷ 32

32 )7104(222

- 64

× 70

- 64

×64

- 64

××

7104 ÷ 32 = 222

3) 2244 ÷ 51

51)2244(44

- 204

×244

- 244

×××

2244 ÷ 51 = 33

4) 1584 ÷ 12

12)1584(132

- 12

×38

- 36

×24

- 24

××

1584 ÷ 12 = 132

5) 1467 ÷ 9

9)1467(163

- 9

56

- 54

27

- 27

××

1467 ÷ 9 = 163

6) 2830 ÷ 28

28)2830(101

- 28

××3

- 0

30

- 28

×2

2830 ÷ 28 = 101 with the remainder 2

7) 9090 ÷ 45

45)9090(202

- 90

××9

- 0

90

- 90

××

9090 ÷ 45 = 202

8) 7000 ÷ 62

62)7000(112

- 62

×80

- 62

180

- 125

×55

7000 ÷ 62 = 112 with the remainder 55

9) 1150 ÷ 15

15)1150(76

- 105

×100

- 90

10

1150 ÷ 15 = 76 with the remainder 10

Hope that my answer is right and helpful to you

Answer:

$32.50 = p

Step-by-step explanation:

$150 = 85 = 85 +85 + 2p

65/2 = 2p/2

The answer is 54.

Angle 6 is equal to angle 7 so we can substitute angle 7 in for any equations that require angle six what we would want to do is (13x + 9)°+(5x + 9)°= 180°, then you would combine like terms which would then give you 18x+18=180 after this you want to isolate your variable by subtracting 18 from both sides which would give you 18x=162 then you would divide 162 by 18 to give you x=9 next to find angle 6 you would substitute x in the equation so it would be (5(9)+9) 5 multiplied by 9 then add 9 which would give you 45 + 9 which equals out to 54.

Answer:

17

Step-by-step explanation:

[tex]perimeter = 2(length + breadth) \\ breadth = (perimeter \div 2) - length[/tex]

[tex]72 \div 2 = 36 \\ 36 - 19 = 17[/tex]

Answer:

y is 17 mm

Step-by-step explanation:

perimeter formula is p= 2l +2w

19 is width

19+19 = 38

72-38=34

34/2= 17

17 is length

I hope it is right:)

Answer:

a regular pentagon has 72 degrees

Step-by-step explanation:

Answer:

B is the answer

Step-by-step explanation:

Amount of tax paid = 43-40 = 3

Tax rate = 3/40 = 0.75

0.75 x 100 = 7.5%

Tax rate = 7.5%

Answer:

C and B

Step-by-step explanation:

(6)

[tex]\frac{3x-7y}{8}[/tex] × [tex]\frac{6}{3x-7y}[/tex] ← cancel 3x - 7y on numerator and denominator

= [tex]\frac{1}{8}[/tex] × [tex]\frac{6}{1}[/tex] = [tex]\frac{6}{8}[/tex] = [tex]\frac{3}{4}[/tex] → C

(7)

[tex]\frac{6x}{x^2-9}[/tex] ÷ [tex]\frac{8}{4x-12}[/tex]

Factorise the denominators of both fractions

x² - 9 = x² - 3² = (x - 3)(x + 3) ← difference of squares

4x - 12 ← factor out 4 from each term

= 4(x - 3)

Then rewrite as

[tex]\frac{6x}{(x-3)(x+3)}[/tex] ÷ [tex]\frac{8}{4(x-3)}[/tex] ← cancel 8 and 4 by 4

= [tex]\frac{6x}{(x-3)(x+3)}[/tex] ÷ [tex]\frac{2}{x-3}[/tex]

• leave first fraction, change ÷ to × , turn second fraction ' upside down'

= [tex]\frac{6x}{(x-3)(x+3)}[/tex] × [tex]\frac{x-3}{2}[/tex] ← cancel x - 3 on numerator and denominator

= [tex]\frac{6x}{x+3}[/tex] × [tex]\frac{1}{2}[/tex] ← cancel 2 and 6 on numerator and denominator

= [tex]\frac{3x}{x+3}[/tex] → B

Answer:

y=−11x+77/2

Step-by-step explanation:

The procedure for solving simultaneous linear equations now called Gaussian elimination appears in the ancient Chinese mathematical text Chapter Eight: Rectangular Arrays of The Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen problems, with two to five equations.[4]

Systems of linear equations arose in Europe with the introduction in 1637 by René Descartes of coordinates in geometry. In fact, in this new geometry, now called Cartesian geometry, lines and planes are represented by linear equations, and computing their intersections amounts to solving systems of linear equations.

The first systematic methods for solving linear systems used determinants, first considered by Leibniz in 1693. In 1750, Gabriel Cramer used them for giving explicit solutions of linear systems, now called Cramer's rule. Later, Gauss further described the method of elimination, which was initially listed as an advancement in geodesy.[5]

In 1844 Hermann Grassmann published his "Theory of Extension" which included foundational new topics of what is today called linear algebra. In 1848, James Joseph Sylvester introduced the term matrix, which is Latin for womb.

Linear algebra grew with ideas noted in the complex plane. For instance, two numbers w and z in {\displaystyle \mathbb {C} }\mathbb {C}  have a difference w – z, and the line segments {\displaystyle {\overline {wz}}}{\displaystyle {\overline {wz}}} and {\displaystyle {\overline {0(w-z)}}}{\displaystyle {\overline {0(w-z)}}} are of the same length and direction. The segments are equipollent. The four-dimensional system {\displaystyle \mathbb {H} }\mathbb {H}  of quaternions was started in 1843. The term vector was introduced as v = x i + y j + z k representing a point in space. The quaternion difference p – q also produces a segment equipollent to {\displaystyle {\overline {pq}}.}{\displaystyle {\overline {pq}}.} Other hypercomplex number systems also used the idea of a linear space with a basis.

Arthur Cayley introduced matrix multiplication and the inverse matrix in 1856, making possible the general linear group. The mechanism of group representation became available for describing complex and hypercomplex numbers. Crucially, Cayley used a single letter to denote a matrix, thus treating a matrix as an aggregate object. He also realized the connection between matrices and determinants, and wrote "There would be many things to say about this theory of matrices which should, it seems to me, precede the theory of determinants".[5]

Benjamin Peirce published his Linear Associative Algebra (1872), and his son Charles Sanders Peirce extended the work later.[6]

The telegraph required an explanatory system, and the 1873 publication of A Treatise on Electricity and Magnetism instituted a field theory of forces and required differential geometry for expression. Linear algebra is flat differential geometry and serves in tangent spaces to manifolds. Electromagnetic symmetries of spacetime are expressed by the Lorentz transformations, and much of the history of linear algebra is the history of Lorentz transformations.

The first modern and more precise definition of a vector space was introduced by Peano in 1888;[5] by 1900, a theory of linear transformations of finite-dimensional vector spaces had emerged. Linear algebra took its modern form in the first half of the twentieth century, when many ideas and methods of previous centuries were generalized as abstract algebra. The development of computers led to increased research in efficient algorithms for Gaussian elimination and matrix decompositions, and linear algebra became an essential tool for modelling and simulations.[5]

Vector spaces

Main article: Vector space

Until the 19th century, linear algebra was introduced through systems of linear equations and matrices. In modern mathematics, the presentation through vector spaces is generally preferred, since it is more synthetic, more general (not limited to the finite-dimensional case), and conceptually simpler, although more abstract.

A vector space over a field F (often the field of the real numbers) is a set V equipped with two binary operations satisfying the following axioms. Elements of V are called vectors, and elements of F are called scalars. The first operation, vector addition, takes any two vectors v and w and outputs a third vector v + w. The second operation, scalar multiplication, takes any scalar a and any vector v and outputs a new vector av. The axioms that addition and scalar multiplication must satisfy are the following. (In the list below, u, v and w are arbitrary elements of V, and a and b are arbitrary scalars in the field F.)[7]

Answer:

4

Step-by-step explanation:

Answer:

4 is the answer

good luck