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Andrew [12]
3 years ago
13

Evaluate the expression. 9! - 4!(5!)

Mathematics
1 answer:
Luda [366]3 years ago
3 0

Answer:

360,000

Step-by-step explanation:

The exclamation point is factorial:

9! = 9×8×7×6×5×4×3×2×1 = 362880

5! = 5×4×3×2×1 = 120

4! = 4×3×2×1 = 24

362880 - 120× 24 = 360000

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<h2>Answer:</h2>

D.

<h2>Step-by-step explanation:</h2>

The representation of this problem is shown below. To find the answer, we need to use the distance formula:

The \ \mathbf{distance} \ d \ between \ the \ \mathbf{points} \ (x_{1},y_{1}) \ and \ (x_{2},y_{2}) \ in \ the \ plane \ is:\\ \\ d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

  • The first diagonal is formed by the points A and C
  • The second diagonal is formed by the points B and D

So, for the first diagonal:

A(x_{1},y_{1})=A(-1,10) \\ \\ C(x_{2},y_{2})=C(1,8) \\ \\ \\ d_{1}=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\ \\ d_{1}=\sqrt{[1-(-1)]^2+(8-10)^2}=2\sqrt{2}

For the second diagonal:

B(x_{1},y_{1})=B(-4,5) \\ \\ D(x_{2},y_{2})=D(4,3) \\ \\ \\ d_{2}=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \\ \\ d_{2}=\sqrt{[4-(-4)]^2+(3-5)^2}=2\sqrt{17}

So the diagonals aren't congruent. Are they perpendicular?

\vec{AC}=(1,8)-(-1,10)=(2,-2) \\ \\ \vec{BD}=(4,3)-(-4,5)=(8,-2)

These two vectors will be perpendicular (hence the diagonals will be perpendicular) if and only if the dot product equals zero, so:

(2,-2).(8,-2)=2(8)+(-2)(-2)=20\neq 0

Thus, the diagonals aren't perpendicular. In conclusion:

<h3>D. The diagonals are neither congruent nor perpendicular, which means the quadrilateral is a parallelogram only.</h3>
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