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mezya [45]
4 years ago
6

Is 3 greater than -2

Mathematics
2 answers:
dimaraw [331]4 years ago
3 0
Yes, three is greater than -2
astraxan [27]4 years ago
3 0
Yes three is greater than -2. If you compare these two numbers on a number line than three will always be greater. Positive numbers are always greater than negative numbers
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Negative. (Zero) positive

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A ball will be drawn from a bag containing 20 balls numbered 1 through 20 . ​If each ball is equally likely to be drawn, what is
lianna [129]
25% or 1/4 because 5 is 25 percent of 20 and if you change 25% into a decimal you get .25 which is equivalent to 1/4
7 0
3 years ago
Read 2 more answers
I dont have brain cells :/
Aleks04 [339]

Answer:

the answer would be 2.

Step-by-step explanation:

(9 - (3 / (1/3)) plus 2 equals 2

Hope this helps!

Good luck

8 0
4 years ago
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10 POINTS! Please help me on the attached file! I keep retaking this over and over again and haven't passed! :D
eduard
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Is 7.64 rational or irrational
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Step-by-step explanation:

4 0
3 years ago
Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than to
pickupchik [31]

1) C (9,-3.5)

2) C (17,-1.5)

Step-by-step explanation:

1)

To solve this problem, we must divide the segment AB into 8 equal intervals, and then find the point sitting at 3/8 of the whole segment.

The end points of the segment in this problem are:

A(3,-5)

and

B(19,-1)

This means that the x- and y-coordinates of point C are given by the equations:

x_c=x_a + 3\frac{x_b-x_a}{8}\\y_c=y_a+3\frac{y_b-y_a}{8}

And substituting the values of the coordinates of A and B, we find:

x_c = x_a + 3 \frac{19-3}{8}=3+3\cdot 2 =9\\y_x = y_a + 3 \frac{-1-(-5)}{8}=-5+3\cdot 0.5 =-3.5

2)

In this problem, we want to find the coordinates of point C such that:

\frac{CB}{AC}=\frac{1}{7} (1)

As before, the coordinates of the endpoints of the segment AB are:

A(3,-5)

and

B(19,-1)

We can call the coordinates of point C as follows:

C(x_c,y_c)

To satisfy eq.(1) for the x-coordinate, we have:

\frac{x_b-x_c}{x_c-x_a}=\frac{1}{7}

Therefore, by substitution we find:

\frac{19-x_c}{x_c-3}=\frac{1}{7}\\7(19-x_c)=x_c-3\\8x_c=136 \rightarrow x_c = 17

Similarly on the y-coordinate we find:

\frac{y_b-y_c}{y_c-y_a}=\frac{1}{7}

And solving we get:

\frac{-1-y_c}{y_c-(-5)}=\frac{1}{7}\\7(-1-y_c)=y_c+5\\8y_c=-12 \rightarrow y_c = -1.5

7 0
3 years ago
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