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My name is Ann [436]
2 years ago
5

Which ordered pairs are in the solution set of the system of linear inequalities?

Mathematics
2 answers:
antoniya [11.8K]2 years ago
5 0

Thought Process:

The solution to this can be found through plotting both of these functions and shading each region above or below the lines (as per the greater and less than signs given)

the region that overlaps both of the above shaded region is the solution set for all ordered pairs that satisfy the two inequalities.

Solution:

let's start by plotting

y > -\dfrac{1}{3}x + 2

it's the same as plotting y = -\dfrac{1}{3}x + 2, but '>' sign suggests to shade everything above this line.

<u>side note:</u>

'>' excludes every ordered pair in the line,

'\leq' includes every ordered pair in the line,

coming back to our solution:

now let's plot the other equation

y > 2x + 3

it's the same as plotting  y = 2x + 3, but '<' sign suggests to shade the area below this line.

The solution set is the area that is overlapped by the two shaded regions.

So every ordered pair (or coordinate (x,y)) that lies within this overlapped shaded region, excluding the points that lie on the lines themselves, satisfy the given two inequalities

Basile [38]2 years ago
3 0

Answer:

Your answer is A) - (2, 2), (3, 1), (4, 2)

Step-by-step explanation:

I hope this helps!

- sincerelynini

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Alex Ar [27]
(7,5)(-4,-1)
slope = (-1-5) / (-4-7) = -6/-11 = 6/11

y - y1 = m(x - x1)
slope(m) = 6/11
(-4,-1)...x1 = -4 and y1 = -1
now we sub...pay close attention to ur signs
y - (-1) = 6/11(x - (-4) ...not done yet...
y + 1 = 6/11(x + 4) <===
6 0
2 years ago
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If UW =9x-9 what is UW as units
Vaselesa [24]
UW is always equal to 9x-9 or 9(x-1) whenever x is a real number.  By adding in x, we can get the value.
 
For example, if x was 4, we add in x, giving us 9*4-9=36-9=27 units or
9(4-1)=9(3)=27 units

Hoped this helped in time!


4 0
3 years ago
The following set of coordinates most specifically represents which figure? (−5, 6), (−1, 8), (3, 6), (−1, 4) (6 points) Paralle
Vladimir [108]

Answer:

Rhombus

Step-by-step explanation:

The given points are A(−5, 6), B(−1, 8), C(3, 6), D(−1, 4).

We use the distance formula to find the length of AB.

|AB|=\sqrt{(-1--5)^2+(8-6)^2}

|AB|=\sqrt{16+4}

|AB|=\sqrt{20}

The length of AD is

|AD|=\sqrt{(-1--5)^2+(6-4)^2}

|AD|=\sqrt{16+4}

|AD|=\sqrt{20}

The length of BC is:

|BC|=\sqrt{(-1-3)^2+(8-6)^2}

|BC|=\sqrt{16+4}

|BC|=\sqrt{20}

The length of CD is

|CD|=\sqrt{(-1-3)^2+(6-4)^2}

|CD|=\sqrt{16+4}

|CD|=\sqrt{20}

Since all sides are congruent the quadrilateral could be a rhombus or a square.

Slope of AB=\frac{8-6}{-1--5}=\frac{1}{2}

Slope of BC =\frac{8-6}{-1-3}=-\frac{1}{2}

Since the slopes of the adjacent sides are not negative reciprocals of each other, the quadrilateral cannot be  a square. It is a rhombus

7 0
3 years ago
Read 2 more answers
Twenty seven minus 3/2 of a nuber x is not more than 36. What is the number?
Irina-Kira [14]

Answer:

Step-by-step explanation:

27 - 1.5x ≤ 36

54 - 3x ≤ 72

54 - 72 ≤3x

-18 ≤ 3x

-6 ≤ x

x ≥ -6

3 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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