1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ivahew [28]
3 years ago
7

How much time is it from 3:45pm to 5:15pm

Mathematics
2 answers:
Gre4nikov [31]3 years ago
7 0
1 hour and 30 minutes
Leokris [45]3 years ago
4 0
1 hour and 30 minutes because when you subtract those you will get that answer
You might be interested in
Write a subtraction problem involving regrouping that has ted reading 304 pages. Answer your question
kari74 [83]
How about this:

The book has 712 pages. Ted has already read 408 pages, so he has 304 left to read.
712 - 408 = 304

4 0
3 years ago
PLEASE HELP PLEASE I BEG YOU. CHOOSE ALL THE RIGHT ONES
andre [41]

Answer:

a c e g

Step-by-step explanation:

yur welcume

4 0
3 years ago
Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u
Tamiku [17]

Answer:

Step-by-step explanation:

15. \frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx}   \\

16.  \frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }

=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2}  \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\

=> The clause is correct

17. Do the same (16)

18. \frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx

19.

\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\

=> The clause is correct

20. Do the same (18)

21. Left = \frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA}  } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}

Right = cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA}   \\

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0
3 years ago
Simplify (2 3/5) divided ( -3 3/4)
Ostrovityanka [42]

Flip over the second fraction

2 3/5= 13/5 ( Multiply the whole number with the denominator, 2*5=10 then add the numerator 10+3=13

-3 3/4= -15/4

13/5 * -4/15= -52/75

Answer :-52/75

4 0
3 years ago
The equation tan(559)= 15 can be used to find the length of AC.
Katarina [22]

Answer:

10.5

Step-by-step explanation:

b=15/tan(55 degrees)=10.5

AC=10.5

8 0
3 years ago
Other questions:
  • Suppose y varies as x, and y=16 wen x=8. find y when x=16
    11·1 answer
  • An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From pre
    5·1 answer
  • Brainliest Answer!!!
    14·1 answer
  • PLEASE PLEASE PLEASE HELP TIMEDCan you prove that DE F = HGF Justify your answer. A. Yes, the triangles are congruent by SAS. B.
    9·2 answers
  • Find the unknown angle measure by solving for the given variable
    12·1 answer
  • What is the system of measurement used most often in the United States?
    11·2 answers
  • Mrs. Potter divides the Grade 8 class of 21 students into few equal groups. The number of
    15·1 answer
  • I wanna cry Joseph pls answer this
    5·1 answer
  • Suppose that during one. Of extreme cold, the average daily temperature decrease by 1 1/2°F each day. How many days did it take
    15·1 answer
  • Guys what is 2+2 im struggling pls help
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!