Math Problem Statement

1.18 Find the vector with the initial point P(4,3) and Q(−9,−4) the terminal point.

(2 Marks)

1.19 Given that u=<10,−1> and v=<10,−4> . Calculate |u−v| .

(2 Marks)

1.20 Given the point A has coordinates (2,7) and B has coordinates (4,11) find the vector rightwards arrow for B A of.

(2 Marks)

1.21 Perform the indicated operations and simplify your answer, (c+72)2 .

(3 Marks)

1.22 Simplify the following expression, 5(3t−4)−(t2+1)−4t(t−1) .

(2 Marks)

1.23 Solve for x in the equation cos(x)=−2√2 .

(2 Marks)

1.24 Convert log84096=4 to exponential form.

(2 Marks)

1.25 Given that f(x)=x+2 and g(x)=x2 . Find g∘f .

(2 Marks)

1.26 Given the function, f(x)=x2−5x+38 and g(x)=(x−17) . Find (f−g)(x) .

(2 Marks)

1.27 Use the table below to find (f+g)(−1) .

(2 Marks)

t

-3

-1

1

3

f(t)

12

8

-18

9

g(t)

-2

-7

-14

-9

Solution

Let's solve each of the problems step by step.

1.18 Find the vector with the initial point P(4,3)P(4,3) and Q(9,4)Q(-9,-4) as the terminal point.

To find the vector PQ\overrightarrow{PQ}, subtract the coordinates of the initial point PP from the coordinates of the terminal point QQ: PQ=QxPx,QyPy=94,43=13,7\overrightarrow{PQ} = \langle Q_x - P_x, Q_y - P_y \rangle = \langle -9 - 4, -4 - 3 \rangle = \langle -13, -7 \rangle So, the vector is PQ=13,7\overrightarrow{PQ} = \langle -13, -7 \rangle.

1.19 Given that u=10,1\mathbf{u} = \langle 10, -1 \rangle and v=10,4\mathbf{v} = \langle 10, -4 \rangle, calculate uv|\mathbf{u} - \mathbf{v}|.

First, subtract the vectors u\mathbf{u} and v\mathbf{v}: uv=1010,1(4)=0,3\mathbf{u} - \mathbf{v} = \langle 10 - 10, -1 - (-4) \rangle = \langle 0, 3 \rangle Now, calculate the magnitude uv|\mathbf{u} - \mathbf{v}|: uv=02+32=9=3|\mathbf{u} - \mathbf{v}| = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 So, uv=3|\mathbf{u} - \mathbf{v}| = 3.

1.20 Given that point AA has coordinates (2,7)(2,7) and BB has coordinates (4,11)(4,11), find the vector BA\overrightarrow{BA}.

To find the vector BA\overrightarrow{BA}, subtract the coordinates of BB from those of AA: BA=AxBx,AyBy=24,711=2,4\overrightarrow{BA} = \langle A_x - B_x, A_y - B_y \rangle = \langle 2 - 4, 7 - 11 \rangle = \langle -2, -4 \rangle So, the vector is BA=2,4\overrightarrow{BA} = \langle -2, -4 \rangle.

1.21 Perform the indicated operations and simplify (c+72)2(c + 72)^2.

Expand the expression: (c+72)2=c2+272c+722=c2+144c+5184(c + 72)^2 = c^2 + 2 \cdot 72 \cdot c + 72^2 = c^2 + 144c + 5184 So, the simplified expression is c2+144c+5184c^2 + 144c + 5184.

1.22 Simplify the following expression: 5(3t4)(t2+1)4t(t1)5(3t - 4) - (t^2 + 1) - 4t(t - 1).

First, expand and distribute: 5(3t4)=15t205(3t - 4) = 15t - 20 4t(t1)=4t2+4t-4t(t - 1) = -4t^2 + 4t Now, combine all terms: 15t20(t2+1)4t2+4t=5t2+19t2115t - 20 - (t^2 + 1) - 4t^2 + 4t = -5t^2 + 19t - 21 So, the simplified expression is 5t2+19t21-5t^2 + 19t - 21.

1.23 Solve for xx in the equation cos(x)=22\cos(x) = -\frac{\sqrt{2}}{2}.

The solutions for cos(x)=22\cos(x) = -\frac{\sqrt{2}}{2} are: x=3π4+2kπorx=5π4+2kπfor integer kx = \frac{3\pi}{4} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{4} + 2k\pi \quad \text{for integer } k

1.24 Convert log84096=4\log_8 4096 = 4 to exponential form.

The equivalent exponential form is: 84=40968^4 = 4096

1.25 Given that f(x)=x+2f(x) = x + 2 and g(x)=x2g(x) = x^2, find (gf)(x)(g \circ f)(x).

To find g(f(x))g(f(x)), substitute f(x)f(x) into g(x)g(x): g(f(x))=g(x+2)=(x+2)2=x2+4x+4g(f(x)) = g(x + 2) = (x + 2)^2 = x^2 + 4x + 4

1.26 Given the functions f(x)=x25x+38f(x) = x^2 - 5x + 38 and g(x)=x17g(x) = x - 17, find (fg)(x)(f - g)(x).

Subtract g(x)g(x) from f(x)f(x): (fg)(x)=(x25x+38)(x17)=x26x+55(f - g)(x) = (x^2 - 5x + 38) - (x - 17) = x^2 - 6x + 55

1.27 Use the table below to find (f+g)(1)(f + g)(-1).

tt-3-113
f(t)f(t)128-189
g(t)g(t)-2-7-14-9

To find (f+g)(1)(f + g)(-1), add f(1)f(-1) and g(1)g(-1): f(1)+g(1)=8+(7)=1f(-1) + g(-1) = 8 + (-7) = 1

So, (f+g)(1)=1(f + g)(-1) = 1.

If you have any questions or need further details, feel free to ask!


Here are some related questions:

  1. How do you find the magnitude of a vector?
  2. What are the properties of the dot product of two vectors?
  3. How do you solve trigonometric equations involving cosine?
  4. What is the importance of converting logarithmic expressions to exponential form?
  5. How do you perform function composition with more complex functions?

Tip: When solving vector problems, always double-check the direction and magnitude to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Vectors
Magnitude of Vectors
Vector Operations
Trigonometric Equations
Logarithmic and Exponential Equations
Function Composition

Formulas

Vector subtraction
Magnitude of a vector
Cosine function inverse
Exponential form of logarithm
Function composition

Theorems

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Suitable Grade Level

Grades 10-12