Q: A current is $$5.0\times10^{-6}$$ amps and becomes $$100$$ times smaller. What is the new current?

Divide by $$100$$. $$100=10^2$$ $$5.0\times10^{-6}\div10^2=5.0\times10^{-8}$$ Final answer: $$5.0\times10^{-8}$$

Q: If $$a$$ and $$b$$ are integers and $$\left(\sqrt{a}\times\sqrt[3]{b}\right)^6=64$$, then $$a+b$$ could equal

Rewrite the expression. $$\left(a^{1/2}b^{1/3}\right)^6=a^3b^2$$ So $$a^3b^2=64$$ Factor $$64$$. $$64=2^6$$ Try integer values of $$a$$. If $$a=4$$ then $$a^3=64$$ So $$b^2=1$$ $$b=\pm1$$ This gives $$a+b=4+1=5$$ or $$a+b=4-1=3$$ Both are possible. Final answer: $$3\text{ and }5$$

Question 1

Compute $$(4	imes10^{18})(5	imes10^{25})$$

Accepted Answer

Multiply coefficients. $$4\cdot5=20$$ Add exponents. $$10^{18}\cdot10^{25}=10^{43}$$ So $$20	imes10^{43}$$ Rewrite. $$2.0	imes10^{44}$$ Final answer: $$2.0	imes10^{44}$$

Question 2

A population triples every $$20$$ minutes. At $$4:00$$ p.m. the population is $$810$$. Compare Column A and Column B.Column A: Population at $$3:00$$ p.mColumn B: $$30$$

Accepted Answer

From $$3:00$$ to $$4:00$$ is $$60$$ minutes. Number of intervals: $$\frac{60}{20}=3$$ Work backward. $$810\div3=270$$ $$270\div3=90$$ $$90\div3=30$$ Column A: $$30$$ Final answer: $$	ext{Equal}$$

Question 3

Solve for $$x$$ in $$8^{2x+1}=16^{x+3}$$

Accepted Answer

Rewrite with base $$2$$. $$8=2^3$$ $$16=2^4$$ So $$2^{3(2x+1)}=2^{4(x+3)}$$ $$6x+3=4x+12$$ $$2x=9$$ $$x=\frac{9}{2}$$ $$\frac{9}{2}=4.5$$ Final answer: $$4.5$$

Question 4

A current is $$5.0	imes10^{-6}$$ amps and becomes $$100$$ times smaller. What is the new current?

Accepted Answer

Divide by $$100$$. $$100=10^2$$ $$5.0	imes10^{-6}\div10^2=5.0	imes10^{-8}$$ Final answer: $$5.0	imes10^{-8}$$

Question 5

Which of the following are equal to $$\left(\frac{1}{720}\right)^{-3}$$? Indicate all correct answers.

Accepted Answer

Use the negative exponent rule. $$\left(\frac{1}{720}\right)^{-3}=720^3$$ Check each choice. $$720^3$$ matches directly. $$\frac{720^5}{720^{-2}}=720^{5-(-2)}=720^7$$ does not match. $$72^3\cdot10^3=(72\cdot10)^3=720^3$$ matches. $$\sqrt{720^6}=720^3$$ because $$720$$>$$0$$. Final answer: $$720^3,\ 72^3\cdot10^3,\ \sqrt{720^6}$$

Question 6

If $$\sqrt{\sqrt{8x}}=\sqrt[4]{4x}$$, what is the greatest possible value of $$x$$?

Accepted Answer

Rewrite both sides as fourth roots. $$\sqrt{\sqrt{8x}}=(8x)^{1/4}$$ $$\sqrt[4]{4x}=(4x)^{1/4}$$ So $$(8x)^{1/4}=(4x)^{1/4}$$ Both expressions require $$x\ge0$$ Raise both sides to the fourth power. $$8x=4x$$ $$4x=0$$ $$x=0$$ Final answer: $$0$$

Question 7

If $$a$$ and $$b$$ are integers and $$\left(\sqrt{a}	imes\sqrt[3]{b}ight)^6=64$$, then $$a+b$$ could equal

Accepted Answer

Rewrite the expression. $$\left(a^{1/2}b^{1/3}ight)^6=a^3b^2$$ So $$a^3b^2=64$$ Factor $$64$$. $$64=2^6$$ Try integer values of $$a$$. If $$a=4$$ then $$a^3=64$$ So $$b^2=1$$ $$b=\pm1$$ This gives $$a+b=4+1=5$$ or $$a+b=4-1=3$$ Both are possible. Final answer: $$3	ext{ and }5$$

Quiz 2

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