4 VctA • VctB (Vector dot product)
AVctA15(VECTOR)7(Dot)VctB=
VCT
5 VctA ⋅ VctB (Vector cross product)
AVctA*VctB=
VCT
6 Obtain the absolute values of VctC.
A1w(Abs)VctC)=
VCT
7 Determine the angle formed by VctA and VctB to three decimal places (Fix 3). v
| • | B) |
| • | B) |
|
(cos = | (A | , which becomes = | (A | ) | ||
AB |
|
| ||||
|
| AB |
1N(SETUP)6(Fix)3
A(VctA15(VECTOR)7(Dot)VctB)/
VCT FIX
(1w(Abs)VctA)1w(Abs) VctB))=
VCT FIX
InequalityCalculations(INEQ)
You can use the following procedure to solve a quadratic inequality or cubic inequality.
1. Press Nc1(INEQ) to enter the INEQ Mode.
2. On the menu that appears, select an inequality type.
To select this inequality type: | Press this key: | |
|
|
|
Quadratic inequality | 1(aX2 | + bX + c ) |
Cubic inequality | 2(aX3 | + bX2 + cX + d ) |
3. On the menu that appears, use keys 1through 4to select the inequality symbol type and orientation.
4. Use the Coefficient Editor that appears to input coefficient values.
• To solve x2 + 2x – 3 < 0, for example, input the coefficients a = 1, b = 2, c =