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3 years ago

The point (2, -3) lies in quadrant I II III IV

2 answers:

4 0

It lies in the IV quadrant ....
1st quadrant (+ , +)
2nd ( - , + )
3rd ( - , - )
4th ( + , -)
....
just think u are in the centre and if u go right thats + and if u go up that is also + ... if u go downward and left thats - ....
We go first in x-axis so
For the 1st quadrant to obtain u should go right and up from centre so (+ , + )
And for 4th quadrant u should go right ( 1st x-axis) and then down so (+ and -)

katovenus [111]3 years ago

3 0

The point (2, -3) lies in the IV quadrant since its x value is positive, but its y value is negative.

*If possible, please give me brainliest*

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The set of points P(0, 3), Q(2, 0), R(4, -3) are collinear and the line has a slope of _____.

jok3333 [9.3K]

Answer:

-3/2,collinear.

Step-by-step explanation:

slope of PQ=(0-3)/(2-0)=-3/2

eq. of line PQ is

y-3=-3/2(x-0)

y-3=-3/2 x

or2y-6=-3x

or 3x+2y=6

if R(4,-3) lies on it,then

3(4)+2(-3)=6

12-6=6

or 6=6

which is true.

Hence points P,Q,R are collinear.

and line has aslope=-3/2

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zimovet [89]

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3 years ago

Read 2 more answers

Convert the following unsigned binary number to unsigned decimal.<br><br> 110011.101

tangare [24]

Answer:

51.625

Step-by-step explanation:

Given a unsigned binary number, to calculate the unsigned decimal:

first, starting at dot, you list the positive powers of 2 from right to left beginning in $2^{0}$ that is equal to “1”. Increase by one the exponent in every power until you complete the total quantity of digits from the unsigned binary number. In this case, since dot to the left, the binary number has six digits (110011). That is to say that you get the followings powers: $2^{5}$ $2^{4}$ $2^{3}$ $2^{2}$ $2^{1}$ $2^{0}$ .

Second, do the same from dot to the right but this time, you list the negative powers of 2 from left to right beginning in $2^{-1}$ that is equal $\frac{1}{2}$ = 0.5. so you get: $2^{-1}$ $2^{-2}$ $2^{-3}$

Now, join two parts and you get:

$2^{5}$ $2^{4}$ $2^{3}$ $2^{2}$ $2^{1}$ $2^{0}$ $2^{-1}$ $2^{-2}$ $2^{-3}$

1 1 0 0 1 1. 1 0 1

Then, you write the equivalent of each of the power below from corresponding binary digit, like that:

$2^{5}$ $2^{4}$ $2^{3}$ $2^{2}$ $2^{1}$ $2^{0}$ $2^{-1}$ $2^{-2}$ $2^{-3}$

1 1 0 0 1 1. 1 0 1

| | | | | | | | |

32 16 8 4 2 1 0.5 0.25 0.125

Finally, you write under the line the equivalent of each power that corresponding to “1” and write “0” under the line, the one that corresponding to “0”, and you sum each one of finals values. Like that:

$2^{5}$ $2^{4}$ $2^{3}$ $2^{2}$ $2^{1}$ $2^{0}$ $2^{-1}$ $2^{-2}$ $2^{-3}$

1 1 0 0 1 1. 1 0 1

| | | | | | | | |

32 16 8 4 2 1 0.5 0.25 0.125

_______________________________________

32 16 0 0 2 1 0.5 0 0.125

32 + 16 + 0 + 0 + 2 + 1 + 0.5+ 0 + 0.125 = 51.625

So that the equivalent from unsigned binary number 110011.101 to unsigned decimal is 51.625

5 0

3 years ago