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Yuri [45]
3 years ago
8

Find the average value of f over region

Mathematics
1 answer:
yan [13]3 years ago
3 0
The area of D is given by:

\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx  \\  \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}

The average value of f over D is given by:

\frac{1}{ \frac{343}{3} }  \int\limits^7_0  \int\limits^{x^2}_0 {4x\sin(y)} \, dydx  = -\frac{3}{343}  \int\limits^7_0 {4x\cos(x^2)} \, dx  \\  \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49
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2 years ago
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9th grade math help me please pretty please
SashulF [63]

Answer:

The system has "infinitely many solutions; consistent and dependent" ⇒ D

Step-by-step explanation:

A consistent system of equations has at least one solution.

  • The consistent independent system has exactly 1 solution
  • The consistent dependent system has infinitely many solutions

An inconsistent system has no solution.

In the system of equations ax + by = c and dx + ey = f, if

  1. a = d, b = e, and c = f, then the system is consistent dependent and has infinitely many solutions
  2. a = d, b = e, and c ≠ f, then the system is inconsistent and has no solution
  3. a ≠ d, and/or b ≠ e, and/or c ≠ f, then the system is consistent independent and has exactly one solution

In the given system of equations

∵ r = -5s + 7

∵ r + 5s - 7 = 0

→ Put the equations in the form of equations above

∵ r = -5s + 7

→ Add -5s to both sides

∴ r + 5s = -5s + 5s + 7

∴ r + 5s = 7 ⇒ (1)

∵ r + 5s - 7 = 0

→ Add 7 to both sides

∴ r + 5s - 7 + 7 = 0 + 7

∴ r + 5s = 7

∴ r + 5s = 7 ⇒ (2)

→ By subtracting equations (1) and (2)

∵ (r - r) + (5s - 5s) = (7 - 7)

∴ 0 + 0 = 0

∴ 0 = 0

→ By using rule 1 above

∵ r = r

∵ 5s = 5s

∵ 7 = 7

∴ The system of equation is consistent dependent and has infinitely

   many solutions

∴ The system has "infinitely many solutions; consistent and dependent"

5 0
2 years ago
Lennox concluded: "It's not possible to map \triangle DCE△DCEtriangle, D, C, E onto \triangle ABE△ABEtriangle, A, B, E using a s
miss Akunina [59]

Answer:

Lennox was curious if triangles \triangle ABE△ABEtriangle, A, B, E and \triangle DCE△DCEtriangle, D, C, E were similar, so she tried to map one figure onto the other using a reflection and a dilation. Lennox concluded: "It's not possible to map \triangle DCE△DCEtriangle, D, C, E onto \triangle ABE△ABEtriangle, A, B, E using a sequence of rigid transformations and dilations, so the triangles are not similar." What error did Lennox make in her conclusion?

Step-by-step explanation:

A rigid transformation is also called an isometry. The transformation of the plane has preserved the size of the triangle (object). So after transformation, the triangle size does not change.

A dilation is an enlargement or reduction of a triangle (object) by a scale factor and with a center of dilation. The scale factor refers to the change in size.

A dilation is a transformation that produces a triangle (object) that is the same shape as the original but is a different size. After dilation, the pre-triangle (pre-object) and triangle (object) have the same shape but not the same size.

Similar figures (object) means figures (object) that have the same shape but may have different sizes.

The figure is not similar, we can not map triangle DCE onto the triangle ABE, both shapes are different.

Line CD can not dilate to line AB with same scale factor.

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