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Licemer1 [7]
3 years ago
15

2.548 rounded to the number 2.55 which place value did Marco round to

Mathematics
2 answers:
JulijaS [17]3 years ago
7 0
The tens place because 4 and 8, 8 is over 5 so u round up then it is 254 rounded to the tens 55
spayn [35]3 years ago
4 0
Marco rounded the thousandths place, or the number 8. Rounding this number will end up with 2.55.

If you round the hundredths, you will get 2.5.
If you round the tenths, you will get 3.
If you round the ones, you will get 0.
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A bag contains white marbles and green marbles, 62 in total. The number of white marbles is 3 less than 4 times the number of gr
Sedaia [141]

Answer:

There are 49 white marbles.

Step-by-step explanation:

w + g = 62        (w = white and g = green marbles).

w = 4g - 3

w - 4g = -3        Subtract this equation from the first equation:

g - (-4g) =  62 - (-3)

5g = 65

g = 65/5 = 13.

So w = 62 - 13 = 49.

3 0
3 years ago
Simplify
melomori [17]

Answer: 4\sqrt{3}

Step-by-step explanation:

For this exercise it is important to remember the following:

i=\sqrt{-1} \\\\i^2=-1

Given the following expression:

-2i\sqrt{-12}

You can notice that the radicand (the number inside the square root) is negative. Therefore, in order to simplify the expression, you need to follow these steps:

1. Replace \sqrt{-1} with i and simplify:

(-2i)(i)\sqrt{12}=-2i^2\sqrt{12}=-2(-1)\sqrt{12}=2\sqrt{12}

2. Descompose 12 into its prime factors:

12=2*2*3=2^2*3

3. Substitute into the expression:

=2\sqrt{2^2*3}

4. Since \sqrt[n]{a^n}=a, you can simplify it:

=(2)(2)\sqrt{3}=4\sqrt{3}

4 0
3 years ago
The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room
tangare [24]
I'm going to assume that the room is a rectangle.

The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle. 

You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
A = lw\\
600 = (x+5)(x+4)\\
600 =  x^{2} + 9x + 20


Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
600 = x^{2} + 9x + 20\\
x^{2} + 9x - 580 = 0\\
(x + 29)(x - 20) = 0\\
x + 29 = 0, \:\: x - 20 = 0\\
x = -29, x = 20

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.

The dimensions of your room are 25ft (length) by 24ft (width).
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3 years ago
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