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3 years ago

The standard unit of measurement for length is the: meter mile yard centimeter

Mathematics

1 answer:

Degger [83]3 years ago

8 0

The meter is the SI unit of measurement for length.

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Diego is planning to make a wooden cookbook stand for his mom. The depth of the stand will be 10 inches and will hold a cookbook

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Answer:

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Step-by-step explanation:

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3 years ago

12000 corn seeds and 92% grew how many plants. How many plants grew. Please explain your reasoning.

Vedmedyk [2.9K]

It's 11040 plants because 92 percent of 12000 is 11040. (Idk if this is the best reasoning but :/)

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3 years ago

Jillian’s school is selling tickets for a play. The tickets cost $10.50 for adults and $3.75 for students. The ticket sales for

Hatshy [7]

Answer:

168 adult tickets

Step-by-step explanation:

3.75(82) (82 students)

307.5

2071.5 - 307.5 ( you subtract since you only need to know the number of adults)

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1764/10.5 (you divide since each adult is 10.5$)

168

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3 years ago

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1.67 ÷ 9.11 = 0.183315038 I need to know the answer simplified.

geniusboy [140]

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3 years ago

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Help! Prove the equality arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

Proof in the explanation

Step-by-step explanation:

Trigonometric Equalities

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

3 years ago