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3 years ago

I need help with this

Mathematics

1 answer:

AleksAgata [21]3 years ago

5 0

I don't see the problem sorry

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PLEASE HELP!!!<br><br> CHOOSE THE SOLUTION TO FOLLOWING INEQUALITY (in picture)

Helga [31]

Answer:

the last one is the answer of it

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3 years ago

Linear Function

monitta

B because x is the independent variable; you could choose any number for x

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3 years ago

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A plane is travelling at 180 metres per second. How many minutes will it take for the plane to travel 800km? Give your answer in

Pavel [41]

Answer:

79 minutes

Step-by-step explanation:

A plane is travelling at 180 meters per seconds

Convert to km/hr

180× 18/5

= 648 km/hr

Since 648 km equals 1 hour then the number of minutes required for the plane to travel 800 km is

= 1/648 × 800

= 1 hour 19 minutes

Conver the time to full minutes

60 minutes = 1 hour

= 60+19

= 79 minutes

Hence the minutes required for the plane to travel 800km is 79 minutes

6 0

3 years ago

40 points!!!!!PLEASE HELP!!!!

Alex17521 [72]

Answer: $\frac{\sqrt{26}}{26}$

This is the square root of 26 over 26. The second 26 is not inside the square root. You can type it in as sqrt(26)/26.

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$\theta$ is in Q1 so $\sin(\theta) > 0$

Use the trig identity below and plug in the given value to get...

$\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}\\\\\sin\left(\frac{2\theta}{2}\right) = \pm \sqrt{\frac{1-\cos(2\theta)}{2}}\\\\\sin(\theta) = \sqrt{\frac{1-\cos(2\theta)}{2}} \ \ \text{since } \ \sin(\theta) > 0\\\\\sin(\theta) = \sqrt{\frac{1-\frac{12}{13}}{2}}\\\\$

$\sin(\theta) = \sqrt{\frac{1}{26}}\\\\\sin(\theta) = \frac{\sqrt{1}}{\sqrt{26}}\\\\\sin(\theta) = \frac{1}{\sqrt{26}}\\\\\sin(\theta) = \frac{1*\sqrt{26}}{\sqrt{26}*\sqrt{26}} \ \ \text{ rationalizing denominator}\\\\\sin(\theta) = \frac{\sqrt{26}}{26}\\\\$

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Edit:

To find cos(theta), we use the pythagorean identity below

$\sin^2(\theta) + \cos^2(\theta) = 1\\\\\cos^2(\theta) = 1-\sin^2(\theta)\\\\\cos(\theta) = \sqrt{1-\sin^2(\theta)}\ \ \text{cosine is positive in Q1}\\\\\cos(\theta) = \sqrt{1-\left(\frac{1}{\sqrt{26}}\right)^2}\\\\\cos(\theta) = \sqrt{1-\frac{1}{26}}\\\\\cos(\theta) = \sqrt{\frac{25}{26}}\\\\\cos(\theta) = \frac{\sqrt{25}}{\sqrt{26}}\\\\\cos(\theta) = \frac{5}{\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{\sqrt{26}*\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{26}\\\\$

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3 years ago

Read 2 more answers

A student formed a pattern in which each term is represented by a sum. The first four terms of the pattern are shown below

Rainbow [258]

N^2 should be the correct answer. Based on those numbers, it appears that the sum can be found by squaring the term n

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3 years ago