Answer:
ai) n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) see attachment
bi) n(C⋂G) = x = 1
bii) n(G) only = 3
Step-by-step explanation:
Let chemistry = C
Economic = E
Government = G
n(E) = 12
n(G) = 8
n(C) = 7
ai) number of pupils for economics and chemistry = 0
number of pupils for economics and government = 4
The set notation for both:
n(E⋂C) = ∅ = null
n(E⋂G) = 4
aii) find attached the Venn diagram
bi) n(C⋂G) = ?
Let number of n(C⋂G) = x
From the Venn diagram
n(C) only = 12-4 = 8
n(G) only = 8-(4+x) = 4-x
n(E) only = 7-x
n(E⋂C⋂G) = 0
n(E⋂C) = 0
n(E⋂G) = 4
Total: 8+ 4-x + 7-x + x + 0+0+4 = 22
23 -x = 22
23-22 = x
x = 1
n(C⋂G) = x = 1
Number of pupils that take both chemistry and government = 1
(bii) government only = n(G) only = 4-x
n(G) only = 4-1 = 3
Number of students that take government only = 3
since there is always a decimal at the end of every value so :
= 262144.0
you move the decimal backwards if there is no decimal between the value. you place the decimal after the 1st digit, you will also count the digits you have passed :
= 2.62144 X 10^5
you passed 5 values so you put 5 in exponent form.
The answer is C.................................................
Answer:
The answer to your question is below
Step-by-step explanation:
Just cancel the parentheses and simplify
Result Addition
5xy² + 2x²y ⇒ (5xy² - x²y) + 3x²y = 5xy² + 2x²y
-3x²y - 2xy² ⇒ x²y + (-2xy² - 4x²y) = -3x²y - 2xy²
3x²y - 2xy² ⇒ 2x²y + (x²y - 2xy²) = 3x²y - 2xy²
5x²y + xy² ⇒ 4x²y + (x²y + xy²) = 5x²y + xy²