Answer:
<em><u>7</u></em><em><u>/</u></em><em><u> </u></em><em><u>Ans;</u></em>

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<em><u>9</u></em><em><u>/</u></em><em><u>Ans;</u></em>

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In the two questions, we first replace the division with multiplication with flipping the fraction after the division sign, secondly we analyze any equation that needs analysis to simplify it, thirdly, to simplify the fraction by deleting the numerator and the similar denominator .
The expression is
5x + 5y
We are to prove that it is an odd integer when x and y are integers of opposite parity
First, we can assume
x = 2a (even)
y = 2b + 1(odd)
subsituting
10(a + b) + 5
5 [(2(a + b) + 1]
The term
2(a + b) + 1 is odd and the result of an odd number multiplied by an odd number is odd
Answer:
Step-by-step explanation:
P(x) = 
Q(x) = 
P(x) × Q(x) = 
= 
= 
P(x) ÷ Q(x) =
÷ 
= 
= 
Given J(1, 1), K(3, 1), L(3, -4), and M(1, -4) and that J'(-1, 5), K'(1, 5), L'(1, 0), and M'(-1, 0). What is the rule that tran
anastassius [24]
(x; y) -> (x - 2; y + 4)
J(1; 1) ⇒ J'(1 - 2; 1 + 4) = (-1; 5)
K(3; 1) ⇒ K'(3 - 2; 1 + 4) = (1; 5)
L(3;-4) ⇒ L'(3 - 2; -4 + 4) = (1; 0)
M(1;-4) ⇒ M'(1 - 2;-4 + 4) = (-1; 0)
Ages of Dan and Cary are 29 and 33 respectively.
<u>Step-by-step explanation:</u>
Step 1:
Form equations out of the given details. Let the age of Dan be x, then age of Cary is x + 4.
In 7 years, sum of their ages = 76
⇒ (x + 7) + (x + 4 + 7) = 76
⇒ 2x + 18 = 76
⇒ 2x = 58
⇒ x = 29
Step 2:
Calculate age of Cary.
⇒ x + 4 = 33