If vector u = <-3.5, -1.5> and vector v = <-1.25, 2.25>, which graph shows the resulting vector for 2v − u?

1 answer:

8 0

Answer:

Step-by-step explanation:

We know that vector addition is scalar addition.

Given vectors are u = <-3.5, -1.5> and v = <-1.25, 2.25>

2v = < -2.5, 4.5>

2v - u = < -2.5, 4.5> - <-3.5, -1.5> = < -2.5 - (-3.5) , 4.5 -(-1.5) >

2v-u = < 1 , 6 >

This vector is basically i + 6j which is drawn in option D

You might be interested in

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$