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aivan3 [116]
3 years ago
9

Estimate the quotient: 1628 ÷ 8

Mathematics
2 answers:
Angelina_Jolie [31]3 years ago
6 0

Answer:

Answer is: 203.5

Step-by-step explanation:

Hope this helped YOU! :)

Pachacha [2.7K]3 years ago
4 0

Answer:

203.5

Step-by-step explanation:

Please mark brainliest.

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help me please will give 11 points yet i know thats a little bit but pls help or i will tell my mommmmyyyy
Andrews [41]

You and your 2 nieces bakes 27 cupcakes. This means that there were 27 cupcakes to be distributed evenly between the 3 of you. So divide 27 by 3 to get your answer.

27 ÷ 3 =  9.

This means that each of you would get 9 miniture cupcakes to take home.

8 0
3 years ago
Find the product. Write your answer in exponential form. 6^-6x6^-1
Tom [10]
So, the questions is 6^(-6) * 6^(-1).

Remember, whenever we are multiplying two numbers with the same base and with different exponents, we should always add the exponents and keeps the same base.

In this ,case it is, -6 and -1.

-6 + (-1) = -7

So the answer would be 6^(-7).

If you have any questions, please comment below.
7 0
3 years ago
What is the Surface area of a cylinder with a radius of 5in and height of 4in
9966 [12]

Answer:

282.74in 2

Step-by-step explanation:

r Radius = 5in

h Height = 4in

3 0
3 years ago
Mary wants to find an expression she can use to determine how far she has traveled. She knows that her distance is represented b
Ivahew [28]

Answer: this one is confusing

Step-by-step explanation:

3 0
2 years ago
Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

Answer:1002

Step-by-step explanation:

\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )

and 1+\tan ^2\theta =\sec^2\theta

and \cos \theta =\frac{1}{\sec \theta }  

\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )

as \cos ^2\theta =\cos ^2(\pi -\theta )

Applying this we get

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]

every \thetathere exist \pi -\theta

such that \sin^2\theta +\cos^2\theta =1

therefore

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002                          

6 0
3 years ago
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