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e-lub [12.9K]
3 years ago
15

Which of the following are identities?

Mathematics
2 answers:
777dan777 [17]3 years ago
4 0

Answer:

all of them

Step-by-step explanation:

work out + simplify the left side and you will get the same as the right side

Nuetrik [128]3 years ago
4 0

Answer:

The answer is 2 because they are similar it is just that the one at the side is factorised

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Students from Williams Middle School are also recycling aluminum cans. The data for the numbers of cans brought in each school d
geniusboy [140]

Answer:

The median would be 18,not enough info for the Interquartile range, and range is 10 to 28

Step-by-step explanation:

You all ready said the answers but, the interquartile range.

6 1
2 years ago
Read 2 more answers
What is the inequality
RideAnS [48]
Is what is an inequality your question? An inequality is basically expressions that use the "greater than", "less than", "greater than or equal to", or "less than or equal to" symbols. A.K.A these:  > <   ≥  ≤
7 0
3 years ago
PLEASE HELP!!!!
pogonyaev

Answer:

30.667, in 1 hour = 40 miles

Step-by-step explanation:

Set up a proportion so 1.5 miles / 2.25 minutes.

So 1.5/2.25 = x/46

solve for x 1.5(46)/2.25 = x so x = 30.667

In one hour is 60 minutes so  1.5(60)/2.25 = x

8 0
2 years ago
Which of the following equations describe the line shown below? <br><br>Check all that apply.
WITCHER [35]
Here, first we need to calculate the slope of the line, 
Here, Coordinates = (-6, 2) (-1, -4)
m = y2 - y1 / x2-x1
m = 2 + 4 / -6 + 1
m = 6/-5
m = -6/5

Now, Take first coordinate: y - 2 = -6/5 (x + 6)
Second coordinate: y + 4 = -6/5 (x + 1)

In short, Your Answers would be Option C & D

Hope this helps!
6 0
3 years ago
 Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III
vodka [1.7K]

Given:

\sin x=-\dfrac{15}{17}

x lies in the III quadrant.

To find:

The values of \sin 2x, \cos 2x, \tan 2x.

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others  are negative.

We know that,

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

(-\dfrac{15}{17})^2+\cos^2 x=1

\cos^2 x=1-\dfrac{225}{289}

\cos x=\pm\sqrt{\dfrac{289-225}{289}}

x lies in the III quadrant. So,

\cos x=-\sqrt{\dfrac{64}{289}}

\cos x=-\dfrac{8}{17}

Now,

\sin 2x=2\sin x\cos x

\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})

\sin 2x=-\dfrac{240}{289}

And,

\cos 2x=1-2\sin^2x

\cos 2x=1-2(-\dfrac{15}{17})^2

\cos 2x=1-2(\dfrac{225}{289})

\cos 2x=\dfrac{289-450}{289}

\cos 2x=-\dfrac{161}{289}

We know that,

\tan 2x=\dfrac{\sin 2x}{\cos 2x}

\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}

\tan 2x=\dfrac{240}{161}

Therefore, the required values are \sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}.

7 0
2 years ago
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