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sammy [17]
3 years ago
12

"Which number can be inserted in the parentheses so the numbers are ordered from least to greatest?" -3,(),-1 1/8 A. -3 1/2 B. 0

C. -2 1/4 D 1 1/2
Mathematics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

Option C

Step-by-step explanation:

The first number is - 3, then we have a blank and the third number is - 1 1/8

In order for the numbers to be arranged from least to greatest, the number in the center should be greater than -3, and lesser than -1 1/8

Note that for negative numbers, the larger the constant, the smaller the number. i.e. -5 is smaller than -4.

So from the given options, the only number that is greater than -3 and lesser than -1 1/8 is - 2 1/4

So, option C gives us the correct answer to have the numbers ordered from least to greatest.

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18/34 as a reduced fraction
olya-2409 [2.1K]

I like to reduce everything by half if I don't already know the answer.

\frac{18}{34}  =  \frac{9}{17}

It just so happens that is the reduced fraction.

All I did was halve 18, which is 9, and halve 34, which is 17.

3 0
3 years ago
Write an equation to represent this line in point slope and slope intercept form....need help please..thanks.
GrogVix [38]
We can use rise over run to determine the slope is 5/8. Using this slope and the point (5,2) we can write the equation: y-2 = 5/8 (x-5) 

Decimal form of 5/8 is .625 , so it could be written y-2 = .625*(x-5)
5 0
4 years ago
Can you answer those 2 questions please?!thank you
Brrunno [24]

Answer:

Question 1: Fraction

Question 2: Equal

Step-by-step explanation:

I know Q1 is right, I'm not so sure about Q2 so I'm sorry if that's wrong

5 0
3 years ago
HELP FAST NOT SURE WHAT TO DO
meriva

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3 0
3 years ago
What is the following sum 3b^2
densk [106]

The given expression is 3b^2*(\sqrt[3]{54a}) + 3*(\sqrt[3]{2ab^6})

This can be simplified as :

= 3*b^2*(\sqrt[3]{27 *2*a}) + 3*(\sqrt[3]{2*a*b^6})

We know that: \sqrt[3]{27}  = 3

Similarly we also can simplify: \sqrt[3]{b^6}  = b^2

So our expression will look like this:

= 3*3*b^2*(\sqrt[3]{2a}) + 3*b^2*(\sqrt[3]{2a})

= 9b^2*(\sqrt[3]{2a}) + 3b^2*(\sqrt[3]{2a})

=\sqrt[3]{2a}*(9b^2 + 3b^2)

=\sqrt[3]{2a}*(12b^2)

This can also be written as:

12b^2(\sqrt[3]{2a})

So the Answer is Option B


6 0
3 years ago
Read 2 more answers
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