Solve the following math problems: 1. Find the value of $(\frac{1}{81})^{-\frac{3}{4}}$. 2. Simplify $\sqrt[3]{27t^{27}}$. 3. Expand and simplify $(2p + 3)(3p - 2)$. 4. y is direct... Solve the following math problems: 1. Find the value of $(\frac{1}{81})^{-\frac{3}{4}}$. 2. Simplify $\sqrt[3]{27t^{27}}$. 3. Expand and simplify $(2p + 3)(3p - 2)$. 4. y is directly proportional to $(x-1)^2$. When x = 3, y = 24. Find y when x = 6.
Understand the Problem
The question contains three math problems:
- Evaluate the expression $(\frac{1}{81})^{-\frac{3}{4}}$.
- Simplify the expression $\sqrt[3]{27t^{27}}$.
- Expand and simplify the expression $(2p + 3)(3p - 2)$.
- Given that $y$ is directly proportional to $(x-1)^2$, and $y=24$ when $x=3$, find the value of $y$ when $x=6$.
Answer
17. (a) $27$ 18. (b) $3t^9$ 19. $6p^2 + 5p - 6$ 20. $y = 150$
Answer for screen readers
- (a) 27
- $3t^9$
- $6p^2 + 5p - 6$
- $y = 150$
Steps to Solve
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Evaluate $(\frac{1}{81})^{-\frac{3}{4}}$ First, we can rewrite the expression using the property $a^{-n} = \frac{1}{a^n}$. $$(\frac{1}{81})^{-\frac{3}{4}} = (81)^{\frac{3}{4}}$$ Next, express 81 as a power of 3: $81 = 3^4$ $$(3^4)^{\frac{3}{4}}$$ Applying the power of a power rule $(a^m)^n = a^{mn}$: $$3^{4 \cdot \frac{3}{4}} = 3^3$$ $$3^3 = 27$$
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Simplify $\sqrt[3]{27t^{27}}$ Rewrite the cube root as a fractional exponent: $$\sqrt[3]{27t^{27}} = (27t^{27})^{\frac{1}{3}}$$ Apply the power to both terms inside the parenthesis: $$(27)^{\frac{1}{3}} \cdot (t^{27})^{\frac{1}{3}}$$ Since $27 = 3^3$, then $27^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = 3^{3 \cdot \frac{1}{3}} = 3^1 = 3$.
Now address the variable: $$(t^{27})^{\frac{1}{3}} = t^{27 \cdot \frac{1}{3}} = t^9$$ Multiply the terms together: $$3 \cdot t^9 = 3t^9$$
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Expand and simplify $(2p+3)(3p-2)$ Expand using the distributive property (FOIL method): $$(2p+3)(3p-2) = (2p)(3p) + (2p)(-2) + (3)(3p) + (3)(-2)$$ $$= 6p^2 -4p + 9p - 6$$ Combine like terms: $$6p^2 + 5p - 6$$
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Find $y$ when $x=6$, given $y$ is directly proportional to $(x-1)^2$, and $y=24$ when $x=3$. Since $y$ is directly proportional to $(x-1)^2$, we can write the relationship as: $$y = k(x-1)^2$$ where $k$ is the constant of proportionality. We are given that $y=24$ when $x=3$. Plug these values into the equation to solve for $k$: $$24 = k(3-1)^2$$ $$24 = k(2)^2$$ $$24 = 4k$$ $$k = \frac{24}{4} = 6$$ So the relationship is: $$y = 6(x-1)^2$$ Now, we want to find $y$ when $x=6$. Substitute $x=6$ into the equation: $$y = 6(6-1)^2$$ $$y = 6(5)^2$$ $$y = 6(25) = 150$$
- (a) 27
- $3t^9$
- $6p^2 + 5p - 6$
- $y = 150$
More Information
Direct proportionality means that one variable is a constant multiple of another. In this case $y$ is proportional to the square of $(x-1)$. The square root of a number $a$, denoted as $\sqrt{a}$, is a value that, when multiplied by itself, gives $a$.
Tips
- Forgetting that a negative exponent means taking the reciprocal and then raising to the power. For example $(\frac{1}{81})^{-\frac{3}{4}}$ is not the same as $(\frac{1}{81})^{\frac{3}{4}}$.
- Not applying fractional exponent rules correctly, e.g. $(a^m)^n = a^{mn}$.
- Incorrectly expanding brackets in the algebra question.
- Not simplifying expressions fully.
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