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Nonamiya [84]
3 years ago
7

PLEASE ANSWET ASAP! -Brainlist to right answer

Mathematics
2 answers:
alexdok [17]3 years ago
8 0
A -

By saying SQRT(50) = (2) X SQRT(25) is wrong because

(2) X SQRT(25) = SQRT(4) X SQRT(25)
= SQRT (100)

B -

To estimate the square room of 50, we have to break 50 into multiples like 1 X 50, 2X25
The ideal multiple should consist at least one integer that can be square root into an integer. Now, let’s start with the multiples of 50.

1 X 50, 2 X 25, 5 X 10,

The most deal multiple is 2 X 25 as it contains an integer 25, that can be squareroot into another integer.
Hence, it will be
SQRT(2) X SQRT(25) =
SQRT (2) X 5 = (1.41) X 5 = 7.05
Flauer [41]3 years ago
4 0

Answer:

aprox 7.1

Step-by-step explanation:

A - it is wrong because he is using wrongly the properties of radicals. He is, nevertheless, using a good intuition.

B

\sqrt[]{50} = \sqrt[]{2\cdot 25} = \sqrt[]{2}\cdot \sqrt[]{25} = 5\sqrt[]{2}

So, since \sqrt[]{2} = 1.41 (aproximately), then

\sqrt[]{50} = 5\cdot 1.41 = 7.07 which is 7.1 when rounded.

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The distance between Taylor’s house and her school is 300,000 centimeters.
Sergio [31]

Answer:

30 km

Step-by-step explanation:

1 kilometer = 100 meters.

1 meter = 100 cm

therefore, 1 km = 10,000

and then, 300000 cm is 30 km

7 0
3 years ago
If you assume that there are exactly 365 days in a year, how many seconds are there in one year? Give your answer to the nearest
Elena L [17]

Answer:

  31,536,000

Step-by-step explanation:

There are 3600 seconds in an hour and 24 hours in a day, so ...

 (3600 s/h)(24 h/da) = 86400 s/da

Then in 365 days, there are ...

  (365 da)(86400 s/da) = 31,536,000 s

_____

No rounding is needed.

5 0
3 years ago
Find the slope of the line passing through the points (-5,6) and (-5,-5)
Leviafan [203]

Hi there! :) :D

Step-by-step explanation:

SLOPE=\frac{Y_2-Y_1}{X_2-X_1}=\frac{RISE}{RUN}

\frac{(-5)-6}{(-5)-(-5)}=\frac{-11}{0}

<u><em>Undefined.</em></u>

<u><em>Therefore, the slope is -11/0</em></u>

<u><em>Final answer is -11/0</em></u>

Hope this helps!

Thanks!

Have a nice day! :)

:D

-Charlie

4 0
3 years ago
How do you solve substitution problems in Pre Algebra
Rus_ich [418]
You multiply both of the equations until you have two of the same term the same like for example, say I have 4x and 5x. You want to multiply until both have the same number, so multiply 5x by four, then multiply 4x by five, and you will get 20x, then both of those cancel out and you will be left with the other variable, and you just solve like a normal equation.
4 0
3 years ago
Read 2 more answers
Subtract.<br> (6x + 5) - (x+3)
Roman55 [17]

(6x+5)-(x+3).

6x+5-x-3.

Add the like terms.

6x-x+5-3.

We get.

5x+2.

6 0
3 years ago
Read 2 more answers
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