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nata0808 [166]
3 years ago
8

What is theformula for finding the hight of a 30-60-90 degree triangle

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
We don't know which side your triangle is sitting on, so the 'height'
could be any one of three values.

Here are the two simple tools that tell you everything you want to
know about a 30°-60°-90° triangle.  These are worth memorizing:

-- The side opposite the 30° angle is 1/2 of the hypotenuse.

-- The side opposite the 60° angle is 1/2 of the hypotenuse times √3 .

I memorized them exactly 60 years ago.  They've been very useful,
and as far as I know, they haven't changed.
You might be interested in
Calculating the degrees of freedom, the sample variance, and the estimated standard error for evaluations using the t statistic
Yuliya22 [10]

Answer:

a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

df = n-1= 10-1=9

d.9

b) s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15

a.15

c) For this case we have the sample size n = 25 and the sample variance is s^2 =400 , the standard error can founded with this formula:

SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80

Step-by-step explanation:

Part a

For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

df = n-1= 10-1=9

d.9

Part b

From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

SS = \sum_{i=1}^n (X_i -\bar X)^2

And the sample variance for this case can be calculated from this formula:

s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15

a.15

Part c

For this case we have the sample size n = 25 and the sample variance is s^2 =400 , the standard error can founded with this formula:

SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80

8 0
3 years ago
What is a 90-degree rotation on a graph.
Oksanka [162]
Rotation of a point through 90-degree is about he origin in clockwise direction when point M(h,k) is rotated about the origin O through 90-degree in clockwise direction
4 0
3 years ago
If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
Ivan

Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

8 0
2 years ago
Find the distance between the points (2, 4) and (8,-8) on a coordinate plane, to the nearest whole number.
Alisiya [41]

Answer:

2,5

Step-by-step explanation:

2 to the right and 5 up

8 0
2 years ago
David is filling out orders for an online business and gets paid $1 for each order he fills out plus bonus of 25 cents per order
svp [43]

Answer:

David will make $481 (he earns the bonus)

Explanation:

<em>If he makes $1 for each order and he filled out 385 orders, then why can't we say he made $385?</em>

Because of this statement rights here:

"...and gets paid $1 for each order he fills out plus bonus of 25 cents per order if the average number of orders he completes per day within any of the given weeks exceeds 20."

So we need to find out if any of the 3 weeks has an average of 20+ orders per day.

<h2>David is filling out orders for an online business and gets paid $1 for each order he fills out</h2>

(x is the amount of orders he fills out)

profit = $1x

<h2>plus bonus of 25 cents per order if the average number of orders he completes per day within any of the given weeks exceeds 20. </h2>

if any average orders per day is > 20 in any week

bonus profit = $1.25x

<h2>The ratio of the number of orders he processed during the first week to the number of orders he processed during the second week is 3:2, </h2>

first week     second week

             3a : 2a

<h2>while the the ratio that compares the number of orders he filled out during the first and the third weeks is 4 to 5 respectively. </h2>

first week   third week

           4a : 5a

<h2>What amount of money will David make at the end of three weeks if the total number of orders he filled out was 385?</h2>

sum of all ratios of a = 385

So we have

3a : <u>2a</u> (first week to <u>second week</u>)

4a : <em>5a </em>(first week to <em>third week</em>)

Notice how the first two numbers are both from the first week. Let's use the Least Common Multiple to make them equal while still keeping ratios.

LCM of 3 and 4: 12 = 3 * 4

12a : <u>8a</u> ( times 4 )

12a : <em>15a</em> ( times 3 )

Now that we have the same value, we can create a big ratio

first week <u>second week</u> <em>third week</em>

   12a     :        <u>8a</u>          :      <em>15a</em>

we know that these ratios will all equal 385. Since ratios are equal no matter how big we make them, we can say that

12a + <u>8</u>a + <em>15</em>a = 385 (a is a variable to scale up the ratio)

which is the same as

(12 + <u>8</u> + <em>15</em>) * a = 385

(<em><u>35</u></em>) * a = 385

35a = 385

if we solve for a by dividing 35 on both sides we get

a = 11

This gives us how much to multiply the RATIO by to get the ACTUAL NUMBER of orders completed. Let's plug 11 for 'a' and see what happens.

12a + <u>8</u>a + <em>15</em>a = 385

12(11) + <u>8</u>(11) + <em>15</em>(11) = 385

132 + <u>88</u> + <em>165</em> = 385     (Check that out, the number of orders each week!)

<u>220</u> + <em>165</em> = 385

<em><u>385</u></em> = 385

Bingo! All the math works out. So, looking back at the verryyy top of this problem, the reason why it wasn't as easy as $385 was because of the bonus.

The bonus gives David $1.25 per order instead of $1 per order if any of the weeks have an average ORDER PER DAY of anything bigger than 20. If we know the real numbers of orders for every week (132, <u>88</u>, and <em>165</em>), then we can divide it by 7 to get the average order per day. Let's choose <em>165 </em>(the <em>third week</em>) because it is the biggest and has the greatest chance of meeting our goal.

165 orders / 7 days (7 days in a week) = 23.57 orders per day

Is this greater than 20 orders per day?

YES!

So now we can safely say that the bonus is there or not, and in this case, the bonus IS there because there is a week where David had more than 20 orders per day.

So instead of using

profit = $1x

We will use

bonus profit = $1.25x

(x is the amount of orders completed)

So if we know he completed 385 orders, and we know he earned the bonus, we plug in 385 for x for the bonus function

bonus profit = $1.25x

bonus profit = $1.25 * 385

bonus profit = $481.25

If necessary, round your answer to the nearest dollar.

So for the very end, all we have to do is round it to the nearest dollar.

$481.25 rounds to $481.

And we're done!

8 0
3 years ago
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