Evaluate 4x^2+3y for x=5 and y=6
2 answers:
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Step-by-step explanation:
<u>Given</u>: √{32⁰ + (2/3)} = (0.6)²⁻³ˣ
<u>Asked</u>: Find the value of x = ?
<u>Solution:</u> Given that √{32⁰ + (2/3)} = (0.6)²⁻³ˣ
⇛√{1 + (2/3)} = (0.6)²⁻³ˣ
⇛√{(1/1) + (2/3) = (0.6)²⁻³ˣ
⇛√{(1*3 + 2*1)/3} = (0.6)²⁻³ˣ
⇛√{(3 + 2)/3} = (0.6)²⁻³ˣ
⇛√(5/3) = (0.6)²⁻³ˣ
Squaring on both sides then
⇛{√(5/3)}² = {(0.6)²⁻³ˣ}²
⇛√(5²/3²) = {0.6}(²⁻³ˣ)²
⇛√{(5*5)/(3*3)} = {0.6}(²⁻³ˣ)²
⇛5/3 = {0.6}(²⁻³ˣ)²
[ (aᵐ)ⁿ = aᵐⁿ]
⇛5/3 = (0.6)²*²⁻³ˣ*²
⇛5/3 = (0.6)⁴⁻⁶ˣ
⇛5/3 = (6/10)⁴⁻⁶ˣ
⇛5/3 = {(6÷2)/(10÷2)})⁴⁻⁶ˣ
⇛5/3 = (3/5)⁴⁻⁶ˣ
⇛(3/5))⁻¹ = (3/5)⁴⁻⁶ˣ
[ a⁻ⁿ = 1/aⁿ]
Base are the same, so the exponents must be equal.
-1 = 4 - 6x
Shift the number 4 from RHS to LHS, changing it's sign.
⇛-1 - 4 = -6x
⇛-5 = -6x
⇛x = {(-5)/(-6)}
x = 5/6
<u>Answer</u><u>:</u> Hence, the value of x for the given problem is 5/6.
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