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4 years ago

What is 45.385 rounded to the nearest hundredth

Mathematics

2 answers:

aliya0001 [1]4 years ago

7 0

Answer:

45.39

Step-by-step explanation:

When the number in the thousandths place is 5 or above (until 9) you round up. If it is 0-4 you round down.

il63 [147K]4 years ago

4 0

Answer:

Your Answer is 45.39 Hope this helps :)

Step-by-step explanation:

Now if you want i can explain it to you

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F(x) =2^x−4<br> g(x) = 5/x<br><br> evaluate g(f(2))=

Afina-wow [57]

Answer:

not possible

Step-by-step explanation:

f(2)=2^2-4=0

g(f(2))=5/0 it's not possible

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3 years ago

A ladder is resting against a wall. The top of the ladder touches the wall at a height of 6

viktelen [127]

Answer:

10 feet

Step-by-step explanation:

let the ladder be x feet in length

Using the Pythagoras theorem,

6^2+(x-2)^2=x^2

if you solve this you'll get, x = 10

Answered by GAUTHMATH

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3 years ago

This table shows the number of girls enrolled in school by class. If a student is chosen at random from this group, which is the

Naya [18.7K]

Answer:

102/526

Step-by-step explanation:

This would be the answer because when u add the amount of freshman plus sophomore plus juinior and plus senior, you get the result of 526. so that would be the denominator while the number of seniors would be the numerator...

6 0

3 years ago

URGENT HELP NEEDED!!!!!

Alona [7]

Answer:

Step-by-step explanation:

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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