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3 years ago

What is the answer to this problem 1/2=q 2/3?

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

6 0

3/2 = 2q
q = 3/2 ÷ 2
q = 3/2 × 1/2
q = 3/4

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Simplified product ?

mel-nik [20]

Answer:

Last choice is correct.

Step-by-step explanation:

$\left(\sqrt{10x^4}-x\sqrt{5x^2}\right)\left(2\sqrt{15x^4}+\sqrt{3x^3}\right)$

$\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)$

$\left(x^2\sqrt{10}-x^2\sqrt{5}\right)\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$x^2\sqrt{10}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)-x^2\sqrt{5}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$2x^4\sqrt{150}+x^3\sqrt{30x}-2\sqrt{75}x^4-x^3\sqrt{15x}$

$2x^4\cdot5\sqrt{6}+x^3\sqrt{30x}-2\cdot5\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10\sqrt{3}x^4-x^3\sqrt{15x}$

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Hence final answer is $10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

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3 years ago

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

german

For this case we have the following function:
$f(x) = tan(\frac{x^2}{9})$
The average rate of change is given by:
$AVR = \frac{f(x2) - f(x1)}{x2- x1}$
Evaluating the function for the given interval we have:
For x = 1.25:
$f(1.25) = tan(\frac{1.25^2}{9})$
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For x = 2:
$f(2) = tan(\frac{2^2}{9})$
$f(2) = 0.48$
Then, replacing values we have:
$AVR = \frac{0.48 - 0.18}{2 - 1.25}$
$AVR = 0.4$
Answer:
the average value of on the closed interval [1.25, 2] is:
$AVR = 0.4$

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3 years ago

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The last answer, because the class is being divided into four groups. So the class number is not known, thus it is represented by “w” and divided by four.

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Step-by-step explanation:

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