Whole numbers<span><span>\greenD{\text{Whole numbers}}Whole numbers</span>start color greenD, W, h, o, l, e, space, n, u, m, b, e, r, s, end color greenD</span> are numbers that do not need to be represented with a fraction or decimal. Also, whole numbers cannot be negative. In other words, whole numbers are the counting numbers and zero.Examples of whole numbers:<span><span>4, 952, 0, 73<span>4,952,0,73</span></span>4, comma, 952, comma, 0, comma, 73</span>Integers<span><span>\blueD{\text{Integers}}Integers</span>start color blueD, I, n, t, e, g, e, r, s, end color blueD</span> are whole numbers and their opposites. Therefore, integers can be negative.Examples of integers:<span><span>12, 0, -9, -810<span>12,0,−9,−810</span></span>12, comma, 0, comma, minus, 9, comma, minus, 810</span>Rational numbers<span><span>\purpleD{\text{Rational numbers}}Rational numbers</span>start color purpleD, R, a, t, i, o, n, a, l, space, n, u, m, b, e, r, s, end color purpleD</span> are numbers that can be expressed as a fraction of two integers.Examples of rational numbers:<span><span>44, 0.\overline{12}, -\dfrac{18}5,\sqrt{36}<span>44,0.<span><span> <span>12</span></span> <span> </span></span>,−<span><span> 5</span> <span> <span>18</span></span><span> </span></span>,<span>√<span><span> <span>36</span></span> <span> </span></span></span></span></span>44, comma, 0, point, start overline, 12, end overline, comma, minus, start fraction, 18, divided by, 5, end fraction, comma, square root of, 36, end square root</span>Irrational numbers<span><span>\maroonD{\text{Irrational numbers}}Irrational numbers</span>start color maroonD, I, r, r, a, t, i, o, n, a, l, space, n, u, m, b, e, r, s, end color maroonD</span> are numbers that cannot be expressed as a fraction of two integers.Examples of irrational numbers:<span><span>-4\pi, \sqrt{3}<span>−4π,<span>√<span><span> 3</span> <span> </span></span></span></span></span>minus, 4, pi, comma, square root of, 3, end square root</span>How are the types of number related?The following diagram shows that all whole numbers are integers, and all integers are rational numbers. Numbers that are not rational are called irrational.
Answer:
k(x) = (3g + 5h)(x) ⇒ (1)
k(x) = (5h - 3g)(x) ⇒ (3)
k(x) = (h - g)(x) ⇒ (2)
k(x) = (g + h)(x) ⇒ (4)
k(x) = (5g + 3h)(x) ⇒ (5)
k(x) = (3h - 5g)(x) ⇒ (6)
Step-by-step explanation:
* To solve this problem we will substitute h(x) and g(x) in k(x) in the
right column to find the corresponding function formula in the
left column
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
- Lets start with the right column
# k(x) = (3g + 5h)(x)
∵ g(x) = -3^x + 5
∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]
- Lets simplify 3 × -3^x
take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)
∴ -3(3 × 3^x) = -(3^x+1)
∴ 3g(x) = -3^x+1 + 15
∵ h(x) = 5 - 3x
∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x
- Now substitute 3g(x) and 5h(x) in k(x)
∵ k(x) = (3g + 5h)(x)
∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify
∴ k(x) = 40 - 3^x+1 - 15x
∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)
* k(x) = (3g + 5h)(x) ⇒ (1)
# k(x) = (5h - 3g)(x)
∵ 5h(x) = 25 - 15x
∵ 3g(x) = -3^x+1 + 15
∵ k(x) = (5h - 3g)(x)
∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify
∴ k(x) = 10 + 3^x+1 - 15x
∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)
* k(x) = (5h - 3g)(x) ⇒ (3)
# k(x) = (h - g)(x)
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
∵ k(x) = (h - g)(x)
∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify
∴ k(x) = 3^x - 3x
∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)
* k(x) = (h - g)(x) ⇒ (2)
# k(x) = (g + h)(x)
∵ h(x) = 5 - 3x
∵ g(x) = -3^x + 5
∵ k(x) = (g + h)(x)
∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify
∴ k(x) = 10 - 3^x - 3x
∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)
* k(x) = (g + h)(x) ⇒ (4)
# k(x) = (5g + 3h)(x)
∵ g(x) = -3^x + 5
∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25
∴ 5g(x) = -5(3^x) + 25
∵ h(x) = 5 - 3x
∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x
- Now substitute 5g(x) and 3h(x) in k(x)
∵ k(x) = (5g + 3h)(x)
∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify
∴ k(x) = 40 - 5(3^x) - 9x
∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)
* k(x) = (5g + 3h)(x) ⇒ (5)
# k(x) = (3h - 5g)(x)
∵ 3h(x) = 15 - 9x
∵ 5g(x) = -5(3^x) + 25
∵ k(x) = (3h - 5g)(x)
∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify
∴ k(x) = 5(3^x) - 9x - 10
∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)
* k(x) = (3h - 5g)(x) ⇒ (6)
