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3 years ago

1/4, 32%, and 0.4 in least to greatest

2 answers:

8 0

It's actually already in least to greatest :) The easiest way to go about this is making it all decimals. 1/4=.25, 32%=.32, 0.4=.40. The order is 1/4, 32%, and 0.4 from least to greatest!

lapo4ka [179]3 years ago

3 0

Put them all in to decimals so dived 1 by 5 and 32 by 100 then they will all be decimals and you can more essay compare them

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12x^2-17xy-5y^2 factorise

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Answer:

(4x + y)(3x-5x)

Step-by-step explanation:

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3 years ago

Choose a number between 67 and 113 that is a multiple of 3, 9, and 12

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3 years ago

Find circumference when pi is 22/7

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3 years ago

What is the solution set to, "Katie wants to collect over 100 seashells. She already has 34 seashells in her collection. Each da

Anastaziya [24]

Answer:

Katie will collect more than 100 seashells in 5 and a half days.

Step-by-step explanation:

Katie already has 34 seashells in her collection.

Each day, she finds 12 more seashells on the beach.

Let x be the number of days Katie is collecting seashells.

In x days, she will collect 12x seashells.

In total, she will collect $34+12x$ seashells.

Katie wants to collect over 100 seashells, so

$34+12x> 100$

Solve this inequality. Subtract 34:

$34+12x-34>100-34\\ \\12x>66\\ \\2x>11\\ \\x>5 \dfrac{1}{2}$

Katie will collect more than 100 seashells in 5 and a half days.

3 0

3 years ago

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

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$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

4 0

3 years ago