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Alika [10]
3 years ago
5

The prism shown has a volume of 798 cm3.

Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

10.5 cm

Step-by-step explanation:

Volume = base area × height

798 = (9.5 × 8) × height

height = 798/76

height = 10.5

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Figure BCDE was transformed using the composition rx-axis ◦ T2,2. If point B on the pre-image lies at (3, 4), what are the coord
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Answer:

The answer is "(5, -6)"

Step-by-step explanation:

Given:

The Pre-image line at point B is: (3, 4)

Solution:

\ rx-axis \circ \  T2,2 (3,4) \\\\\Rightarrow  \ rx-axis \ ( T2,2 (3,4) )\\\\ \Rightarrow \ rx-axis ( 5,6  ) \\\\ \Rightarrow  (5, -6)\\

The coordinates of B point is (5,-6).

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
6 0
3 years ago
The graph of f(x) = 2x is shown on the grid.
solniwko [45]

Answer:

  see below

Step-by-step explanation:

Oddly enough, it is the one that with f(x) <em>reflected over the y-axis</em>. All points on the graph are mirrored across that axis (x is changed to -x, y is left alone).

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