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2 years ago

3. A branch breaks off the top of a giant sequoia tree and falls straight to the ground. The branch's height in

Mathematics

1 answer:

Juliette [100K]2 years ago

4 0

Answer:

Its 4.3

Step-by-step explanation:

If you graph it, y is the height an t is the time. At a height of 0 it had been in the air for 4.3 seconds

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

What is the value of 7/9 x 2/3 ?

Elan Coil [88]

Answer:

42/81 or 0.5185185

Step-by-step explanation:

7/9 × 2/3

The hcf of 9 and 3 is 9

7/9 × 2/3

2/3 × 3= 6/9

7/9× 6/9 = 42/81 or 0.5185185

5 0

3 years ago

Read 2 more answers

The other answer choice is 30

il63 [147K]

Answer:

21 pencils

Step-by-step explanation:

b 21 pencils

8 0

2 years ago

Divide the combinations.<br><br> 60C3 / 15C3

Mariulka [41]

I hope you know what '!' means
2!=2
3!=6
4!=24
basicalkly times every natural number including and before that number

nCr= $\frac{n!}{r!(n-r)!}$
so
$\frac{60C3}{15C3}$ =
$\frac{ \frac{60!}{3!(60-3)!} }{ \frac{15!}{3!(15-3)!} }$ =
$( \frac{60!}{3!(60-3)!})(\frac{3!(15-3)!}{15!})$ =
$( \frac{60!}{3!(57)!})(\frac{3!(12)!}{15!})$ =
$\frac{(60!)(3!)(12!)}{(3!)(57!)(15!)}$ =
$\frac{(60!)(12!)}{(57!)(15!)}$ =
$\frac{(60!)}{(57!)(15)(14)(13)}$ =
$\frac{(60)(59)(58)}{(15)(14)(13)}$ =
$\frac{(30)(59)(58)}{(15)(7)(13)}$ =
$\frac{(6)(59)(58)}{(3)(7)(13)}$ =
$\frac{6844}{91}$

4 0

3 years ago

Read 2 more answers

How do I solve this I'm having a problem.

Llana [10]

How do you solve what

4 0

3 years ago

Read 2 more answers