Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Given the vector v has an initial point at (-1,0) and a terminal point at (2,-3), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−1,0) (-1,0) and a terminal point at (2,−3) (2,-3) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Transformations of absolute value functions: translations and reflectionsright chevron icon

Full solution

Q. Given the vector v \mathbf{v} has an initial point at (−1,0) (-1,0) and a terminal point at (2,−3) (2,-3) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Components of Vector: To find the magnitude of vector v, we need to calculate the difference in the x-coordinates and the y-coordinates of the terminal and initial points to get the components of the vector.\newlineThe x-component of vector v is the difference between the x-coordinate of the terminal point and the x-coordinate of the initial point: xcomponent=2−(−1)=3 x_{component} = 2 - (-1) = 3 .\newlineThe y-component of vector v is the difference between the y-coordinate of the terminal point and the y-coordinate of the initial point: ycomponent=−3−0=−3 y_{component} = -3 - 0 = -3 .
  2. Use Pythagorean Theorem: Now that we have the components of vector v, we can calculate its magnitude using the Pythagorean theorem. The magnitude of vector v, denoted as ∣∣v∣∣ ||v|| , is the square root of the sum of the squares of its components.\newlineSo, ∣∣v∣∣=xcomponent2+ycomponent2 ||v|| = \sqrt{x_{component}^2 + y_{component}^2} .\newlineSubstitute the values we found for the components: ∣∣v∣∣=32+(−3)2 ||v|| = \sqrt{3^2 + (-3)^2} .
  3. Square Components and Add: Perform the squaring of the components and add them together: ∣∣v∣∣=9+9 ||v|| = \sqrt{9 + 9} .
  4. Simplify Square Root: Simplify the expression under the square root: ∣∣v∣∣=18 ||v|| = \sqrt{18} .
  5. Factor Out Perfect Square: We can simplify 18 \sqrt{18} by factoring out the square of a whole number. The number 1818 can be written as 9×2 9 \times 2 , and since 99 is a perfect square, we can take its square root out of the radical.\newlineSo, ∣∣v∣∣=9×2=9×2=32 ||v|| = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} .

More problems from Transformations of absolute value functions: translations and reflections

Question
Find g(x) g(x) , where g(x) g(x) is the translation 2 2 units left of f(x)=∣x∣ f(x) = |x| .\newlineWrite your answer in the form a∣x−h∣+k a|x - h| + k , where a a , h h , and k k are integers.\newlineg(x)= g(x) = ______\newline
Get tutor helpright-arrow

Posted 2 months ago

Question
Find g(x) g(x) , where g(x) g(x) is the reflection across the y y -axis of f(x)=∣x∣ f(x) = |x| .\newlineWrite your answer in the form a∣x−h∣+k a|x - h| + k , where a a , h h , and k k are integers.\newlineg(x)= g(x) = ______\newline
Get tutor helpright-arrow

Posted 2 months ago

Question
The graph of y=∣x∣ y=|x| is reflected across the x x -axis and then scaled vertically by a factor of 77 .\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=7∣x∣ y=7|x| \newline(B) y=−7∣x∣ y=-7|x| \newlineC) y=17∣x∣ y=\frac{1}{7}|x| \newline(D) y=−17∣x∣ y=-\frac{1}{7}|x|
Get tutor helpright-arrow

Posted 2 months ago

Question
The graph of y=∣x∣ y=|x| is scaled vertically by a factor of 66 .\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=∣x−6∣ y=|x-6| \newline(B) y=6∣x∣ y=6|x| \newline(C) y=16∣x∣ y=\frac{1}{6}|x| \newline(D) y=∣x−16∣ y=\left|x-\frac{1}{6}\right|
Get tutor helpright-arrow

Posted 1 month ago

Question
The graph of y=∣x∣ y=|x| is scaled vertically by a factor of 13 \frac{1}{3} .\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=∣x∣−3 y=|x|-3 \newline(B) y=−3∣x∣ y=-3|x| \newline(C) y=13∣x∣ y=\frac{1}{3}|x| \newline(D) y=∣x∣−13 y=|x|-\frac{1}{3}
Get tutor helpright-arrow

Posted 2 months ago

Question
The graph of y=∣x∣ y=|x| is shifted to the left by 66 units.\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=∣x+6∣−6 y=|x+6|-6 \newline(B) y=∣x−6∣−6 y=|x-6|-6 \newline(C) y=∣x∣−6 y=|x|-6 \newline(D) y=∣x+6∣ y=|x+6|
Get tutor helpright-arrow

Posted 1 month ago

Question
The graph of y=∣x∣ y=|x| is reflected across the x x -axis and then scaled vertically by a factor of 14 \frac{1}{4} .\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=−4∣x∣ y=-4|x| \newline(B) y=∣x∣−4 y=|x|-4 \newline(C) y=−14∣x∣ y=-\frac{1}{4}|x| \newline(D) y=∣x−4∣ y=|x-4|
Get tutor helpright-arrow

Posted 2 months ago

Question
The graph of y=∣x∣ y=|x| is shifted down by 99 units.\newlineWhat is the equation of the new graph?\newlineChoose 11 answer:\newline(A) y=∣x∣−9 y=|x|-9 \newline(B) y=∣x+9∣ y=|x+9| \newline(C) y=∣x∣+9 y=|x|+9 \newline(D) y=∣x−9∣ y=|x-9|
Get tutor helpright-arrow

Posted 4 months ago

Question
y=13x+5y=2x \begin{array}{l} y=\frac{1}{3} x+5 \\ y=2 x \end{array} \newlineConsider the given system of equations. If (x,y) (x, y) is the solution to the system, then what is the value of y+x y+x ?
Get tutor helpright-arrow

Posted 1 month ago

Question
13x+3y=4 \frac{1}{3} x+3 y=4 \newline2x−4=2y 2 x-4=2 y \newlineConsider the given system of equations. If (x,y) (x, y) is the solution to the system, then what is the value of x−y x-y ?
Get tutor helpright-arrow

Posted 1 month ago

View all (56)