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igomit [66]
2 years ago
9

Jason and Mateo went to eat at Sam's Sub Shop. They spent $12.25 for lunch plus 8% tax. If they split the bill evenly, what was

the price that each person paid
Mathematics
1 answer:
7nadin3 [17]2 years ago
5 0

Answer:

answer 26

ok now help me

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The width of a rectangle is x. Its length is 3x. How long is the diagonal of the rectangle? Express your answer in simplified ra
nikklg [1K]

Answer:

The length of the diagonal is x√10

Step-by-step explanation:

Here, we want to find the length of the diagonal

The diagonal will represent the hypotenuse of a triangle formed with the width and length of the triangle being the measure of the other sides

Mathematically, we then apply Pythagoras’ theorem to get this

we have this as that the square of the diagonal equals the sum of the squares of the two other sides

d^2 = x^2 + (3x)^2

d^2 = x^2 + 9x^2

d^2 = 10x^2

d = √(10x^2)

d = x√10

7 0
2 years ago
What is the domain of the function (-1,1) (0,0) (1,2)
galben [10]

Answer:

{-1,0,1}

Step-by-step explanation:

The domain of a function is the set of all input values. The input values are the first coordinate in any coordinate pair (x,y).

So the domain here is {-1, 0, 1}.

8 0
2 years ago
Which transformation shows a reflection of ΔDEF?
m_a_m_a [10]

The answer is B or the purple one

7 0
3 years ago
Read 2 more answers
What is the period of f(x)
Stells [14]

Step-by-step explanation:

The period of f(x) is π.

To calculate the period using the formula derived from the basic sine and cosine equations. The period for function y = A sin (B a – c) and y = A cos ( B a – c ) is equal to 2πB radians. The reciprocal of the period of a function is equal to its frequency.

8 0
2 years ago
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If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?
Kipish [7]

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

Step-by-step explanation:

The given is,

                        2 x^{3} + 4 x^{2} -30x................................(1)

Step:1

    Check for option A,

             x = 1, dimensions 8 by 9 by 1  

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(2)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                     = 72............................................(3)

            From equation (2) and (3)

                                -24 ≠ 72

            So, X=1; dimensions 8 by 9 by 1 is not possible.

   Check for option B,

             x = 1, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(4)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30.........................................(5)

            From equation (4) and (5)

                                -24 ≠ 30

            So, X=1; dimensions 2 by 5 by 3 is not possible.

   Check for option C,

            x = 4, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72.............................................(6)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30............................................(7)

            From equation (6) and (7)

                               72 ≠ 30

            So, X=4; dimensions 2 by 5 by 3 is not possible.

    Check for option C,

            x = 4, dimensions 8 by 9 by 1

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72............................................(8)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                    = 72............................................(9)

            From equation (8) and (9)

                               72 = 72

            So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

           The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

         

4 0
3 years ago
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