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Leona [35]
3 years ago
5

If SE = 7 and EG = 20 , then SG =

Mathematics
1 answer:
Rina8888 [55]3 years ago
6 0
Try 13 and if it’s wrong im sorry
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What is the area of the trapezoid shown below?<br> 2<br> 15<br> 12<br><br> help fast please!!!
Ugo [173]

Answer:

78 units2

Step-by-step explanation:

The Area of the triangle is 54 units2

The rectangle on the side has an area of 24 units2.

Add them together, and you get 78.

I hope this helps!

pls ❤ and give brainliest pls

7 0
3 years ago
A coin is tossed. If heads appears, a spinner that can land on any number from 1 to 3 is spun. If tails appears, a second coin i
Advocard [28]

Answer:

H1 H2 H3 TH TT

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Reflection across x=-2​
lara31 [8.8K]

Hello,

answer in the jointed picture

6 0
3 years ago
Read 2 more answers
Is anybody else here to help me ??​
Akimi4 [234]

Answer:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

Step-by-step explanation:

I'm going to use x instead of \theta because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

\cot(x)+\cot(\frac{\pi}{2}-x)

I'm going to use a cofunction identity for the 2nd term.

This is the identity: \tan(x)=\cot(\frac{\pi}{2}-x) I'm going to use there.

\cot(x)+\tan(x)

I'm going to rewrite this in terms of \sin(x) and \cos(x) because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: \frac{\cos(x)}{\sin(x)}=\cot(x) and \frac{\sin(x)}{\cos(x)}=\tan(x)

So we have:

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

I'm going to factor out \frac{1}{\sin(x)} because if I do that I will have the \csc(x) factor I see on the right by the reciprocal identity:

\csc(x)=\frac{1}{\sin(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show \cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)} is equal to \csc(\frac{\pi}{2}-x).

So since I want one term I'm going to write as a single fraction first:

\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}

Find a common denominator which is \cos(x):

\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}

\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}

\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}

By  the Pythagorean Identity \cos^2(x)+\sin^2(x)=1 I can rewrite the top as 1:

\frac{1}{\cos(x)}

By the quotient identity \sec(x)=\frac{1}{\cos(x)}, I can rewrite this as:

\sec(x)

By the cofunction identity \sec(x)=\csc(x)=(\frac{\pi}{2}-x), we have the second factor of the right hand side:

\csc(\frac{\pi}{2}-x)

Let's just do it all together without all the words now:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

7 0
3 years ago
Bill went to the museum at 11:30 A.M . He stayed for 3 1/2 hours.When did he leave ?
tatuchka [14]
To start, note that an hour is 60 minutes long. A 1/2 hour, or half hour, is then 60/2=30 minutes. Therefore, when we have 11 hours and 30 minutes, we have 11 and a half hours. Adding 3 and a half to that, we get 11.5+3.5=15 (a half can also be expressed as .5, although it's not typically done that way when expressing time - it just might be easier to visualize it this way). Therefore, we are 15 hours into the day. However, we can't just stop there - we have to account for AM and PM. Therefore, we subtract 12 hours from 15. If the number is positive, we are in PM - otherwise, we're in AM. Therefore, as 15-12=3, the time is in PM. The remaining number is the time, so Bill leaves at 3 PM. If we are left with a decimal (e.g. 3.25), we would keep the 3 and multiply the 0.25 (the decimal) by 60 to figure out how many minutes we have, so 3.25 would turn into 3+0.25*60=3:15.

Feel free to ask further questions!
5 0
3 years ago
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