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Greeley [361]
3 years ago
14

What is the sign of 3 x y 3xy3, x, y when x > 0 x>0x, is greater than, 0 and y < 0 y<0

Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

The sign is negative

Step-by-step explanation:

Given

Expression = 3xy

x > 0

y

Required

Determine the sign

The expression can be split into 3 factors

(3) (x) (y)

If x > 0 then x is positive and If y < 0 then y is negative

So, the expression becomes:

(+3) (+x) (-y)

<em />

Leave only signs

(+) (+) (-)

+ * + = -

So, we have:

(+*+) (-)

(+) (-)

+ * - = -

So, we have:

(+) (-) = -

<em>Hence, the sign of the expression is negative</em>

<em></em>

Take for instance;

x = 2

y = -4

The expression would be:

3 *2 * - 4 =-24 -- negative

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Please help this is important for me.
Anit [1.1K]
<h3>Answer: point Q is located at (-1, 1)</h3>

=================================================

Explanation:

Check out the diagram below.

Plot R(-3,7) and T(3,-11) on the same xy grid.

Draw a vertical line through R and a horizontal line through T. A right triangle forms. At the intersection point is point S(-3,-11)

----------------------

Now measure the distance from point S to point T. You can count out the spaces or subtract the x coordinates and use absolute value.

|x1-x2| = |-3-3| = |-6| = 6

From point S to point T is 6 units.

We want to subdivide this horizontal length in the ratio 1:2

What this means is that we want to plot a point U somewhere such that SU:UT = 1:2

In other words,

SU = x

UT = 2x

SU+UT = ST

x+2x = 6

3x = 6

x = 6/3

x = 2

So we must move 2 spaces to the right from point S to get to U(-1,-11)

Going from point U(-1,-11) to T(3,-11) is 4 spaces

We have SU:UT = 2:4 = 1:2 to help confirm we have the correct location for point U

From point U, we then move straight up to the line segment RT

We'll land on Q(-1,1)

----------------------

Another way to find the y coordinate of point Q is to subdivide the segment RS into the ratio 1:2 similar to how we divided ST up

Segment RS is 18 units long since we go from y = 7 to y = -11 when going from R to S.

If V was on segment RS such that

RV:VS = 1:2

and RV = y

then

RV+VS = RS

y+2y = 18

3y = 18

y = 6

RV = y = 6

VS = 2y = 2*6 = 12

So you'll move 6 units down from y = 7 to land on y = 1 (when going from the y coordinate of R to the y coordinate of Q)

3 0
3 years ago
What is the answer to this
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Adding (a, b) to the above equation gives
  2(a, b) - (x1, y1) = (x2, y2)
Adding (x1, y1) then gives
  2(a, b) = (x1, y1) + (x2, y2)
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  (a, b) = ((x1, y1) + (x2, y2))/2
The midpoint is the average of the end points.

a. Using the result from part b, the midpoint is
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5 0
3 years ago
Solve the system 6x -2y+z= -2 2x+ 3y - 3z =11 x+ 6y=31
weqwewe [10]

Answer:

x = 1

y = 5

z = 2

Step-by-step explanation:

System of equations:

6x - 2y + z = -2

2x + 3y - 3z = 11

x + 6y = 31

Isolate one variable in any of the equations:

x + 6y = 31

x = 31 - 6y

Plug in this value for x in another equation:

6(31 - 6y) - 2y + z = -2

186 - 36y - 2y + z = -2

186 - 38y + z = -2

-38y + z = -188

z = -188 + 38y

Plug in these values in the remaining equation:

2(31 - 6y) + 3y - 3(-188 + 38y) = 11

62 - 12y + 3y + 564 - 114y = 11

626 - 12y + 3y - 114y = 11

626 - 9y - 114y = 11

626 - 123y = 11

-123y = -615

y = 5

Plug in value of y into our other answers to solve for x and z:

x = 31 - 6(5)

x = 31 - 30

x = 1

z = -188 + 38(5)

z = -188 + 190

z = 2

Check your work:

6x - 2y + z = -2

6(1) - 2(5) + 2 = -2

6 - 10 + 2 = -2

-4 + 2 = -2

-2 = -2

Correct!

*Note there are several ways to solve for these types of problems. I used substitution*

8 0
3 years ago
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