Question
Calculate \(\sqrt{17.8}-1.3^{2.5}\)
Answer/Explanation
2.29 or 2.292
Question
(a) \(3^x = \sqrt[4]{3^5}\)
Find the value of x.
(b) Simplify \((32y^{15})^{\frac{2}{5}}\)
Answer/Explanation
Ans:
(a) \(\frac{5}{4}\) oe
(b) \(4y^6\)
Question
Simplify
(a) $\left(\frac{p^4}{16}\right)^{0.75}$,
(b) $3^2 q^{-3} \div 2^3 q^{-2}$.
▶️Answer/Explanation
(a)$\left(\frac{p^4}{16}\right)^{0.75} = \frac{p^{4 \times 0.75}}{16^{0.75}}$
Simplifying the exponents:
$\frac{p^3}{16^{0.75}}$
To simplify further, we can evaluate the exponent $16^{0.75}$:
$16^{0.75} = \sqrt[4]{16^3} = \sqrt[4]{4096} = 8$
Substituting this back into the expression:
$\frac{p^3}{8}$
Therefore, $\left(\frac{p^4}{16}\right)^{0.75}$ simplifies to $\frac{p^3}{8}$.
(b) To simplify $3^2 q^{-3} \div 2^3 q^{-2}$, we can simplify each term separately and then perform the division:
$3^2 = 9$
$q^{-3} = \frac{1}{q^3}$
$2^3 = 8$
$q^{-2} = \frac{1}{q^2}$
Substituting these simplified terms back into the expression:
$9 \cdot \frac{1}{q^3} \div 8 \cdot \frac{1}{q^2}$
Applying the division rule for exponents (subtracting exponents when dividing):
$\frac{9}{8} \cdot \frac{1}{q^3} \cdot q^2$
Simplifying:
$\frac{9}{8q}$
Therefore, $3^2 q^{-3} \div 2^3 q^{-2}$ simplifies to $\frac{9}{8q}$.