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4 years ago

Umeko and Clemente are simplifying the problem below, who was wrong? Explain the error.

Mathematics

1 answer:

lapo4ka [179]4 years ago

8 0

$\log_76+\log_73-\log_72 \\ =\log_7(6\times 3)-\log_72\\ =\log_718-\log_72 \\=\log_7 \frac{18}{2} \\ =\log_79$

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Write the expression using rational exponents. Then simplify and convert back to radical notation.

ioda

Answer:

The radical notation is $3x\sqrt[3]{y^2z}$

Step-by-step explanation:

Given

$\sqrt[3]{27 x^{3} y^{2} z}$

Step 1 of 1

Write the expression using rational exponents.

$\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}$

$=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}$

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

4 0

3 years ago

Find the perimeter of the figure <br> 13.5 ft<br> 9 ft<br> 12.5 ft

saveliy_v [14]

Answer:

Add all the values up and then divide by 2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers

What is 6.27 divided by 1000?

Effectus [21]

Answer:

0.00627 :)

Step-by-step explanation:

3 0

3 years ago

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Music Jeremy spent $33 on 3 CDs .At this a rate how much would 5 CDs cost

kvasek [131]

Question: Jeremy spent $33 on 3 CDs .At this a rate how much would 5 CDs cost?

Answer: I think the answer is 55$.

Hope this helps. c;

8 0

3 years ago

Read 2 more answers