Answer:
Two consecutive numbers whose squares differ by 33 are 16 and 17
Step-by-step explanation:
lets assume first number be x
since numbers are consecutive , so other number will be x + 1
From given information in question
(x + 1)² - x² = 33
⇒ (x² + 1² + 2x) - x² = 33 [ since (a+b)² = a² + b² + 2ab ]
⇒ x² + 1² + 2x - x² = 33
⇒ 2x + 1 = 33
⇒ 2x = 33 - 1
⇒ x = 32/2 = 16
so one number is x = 16 and other number is x + 1 = 16 + 1 = 17
lets recheck our solution
17² - 16² = 289 - 256 = 33 , And since difference is 33 , two required consecutive numbers are 16 and 17.
Answer:
To solve for an equation with two variables, substitute a zero(0) into the other variable and find one variable.
Step-by-step explanation:
To solve for an equation with two variables, substitute a zero(0) into the other variable and find one variable.
Simple Example:
3x+3y=12
Substitute 0 as y,
3x+0=12
3x=12
12/3=4
x=4
Now substitute 4 into the main equation,
3(4)+3y=12
12+3y=12
-12 -12
3y=0
0/3=0
y=0
So the answer is
3(4)+3(0)=12
12=12
9514 1404 393
Answer:
(3) p^3 -3p
Step-by-step explanation:
(a +1/a) = p . . . . . . . given
(a +1/a)^3 = p^3 . . . . . . . . cube both sides
a^3 +3a^2(1/a) +3a(1/a)^2 +(1/a)^3 = p^3 . . . . . . expand
(a^3 +1/a^3) +3(a +1/a) = p^3 . . . . . . . . . . simplify, group
(a^3 +1/a^3) +3p = p^3 . . . . . . . . . . substitute p for a+1/a
(a^3 +1/a^3) = p^3 -3p . . . . . . subtract 3p from both sides
Answer:
suusy
Step-by-step explanation:
sus
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