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2 years ago

Which graph shows the solution to the system of linear inequalities? y < 2x – 5 y > –3x + 1

Mathematics

2 answers:

Ksenya-84 [330]2 years ago

5 0

Answer:

Where are the graphs

Step-by-step explanation:

crimeas [40]2 years ago

4 0

Answer:

Your answer is graph B

Step-by-step explanation:

(I can't provide the graph)

--This is right on edg2020! I hope this helps!--

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What other points are on the line of direct variation through (5, 12)? Check all that apply. (0, 0) (2.5, 6) (3, 10) (7.5, 18) (

aalyn [17]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form $y/x=k$ or $y=kx$

in this problem we have

the point $(5,12)$ is on the line of direct variation

Find the constant of proportionality k

$y/x=k$ -------> substitute ------> $k=12/5$

the equation is

$y=\frac{12}{5}x$

Remember that

If a point is on the line of direct variation

then

the point must satisfy the equation of direct variation

we're proceeding to verify each point

case A) point $(0,0)$

$x=0\ y=0$

Substitute the value of x and y in the direct variation equation

$0=\frac{12}{5}*0$

$0=0$ -------> is true

therefore

the point $(0,0)$ is on the line of direct variation

case B) point $(2.5,6)$

$x=2.5\ y=6$

Substitute the value of x and y in the direct variation equation

$6=\frac{12}{5}*2.5$

$6=6$ -------> is true

therefore

the point $(2.5,6)$ is on the line of direct variation

case C) point $(3,10)$

$x=3\ y=10$

Substitute the value of x and y in the direct variation equation

$10=\frac{12}{5}*3$

$10=7.2$ -------> is not true

therefore

the point $(3,10)$ is not on the line of direct variation

case D) point $(7.5,18)$

$x=7.5\ y=18$

Substitute the value of x and y in the direct variation equation

$18=\frac{12}{5}*7.5$

$18=18$ -------> is true

therefore

the point $(7.5,18)$ is on the line of direct variation

case E) point $(12.5,24)$

$x=12.5\ y=24$

Substitute the value of x and y in the direct variation equation

$24=\frac{12}{5}*12.5$

$18=30$ -------> is not true

therefore

the point $(12.5,24)$ is not on the line of direct variation

case F) point $(15,36)$

$x=15\ y=36$

Substitute the value of x and y in the direct variation equation

$36=\frac{12}{5}*15$

$36=36$ -------> is true

therefore

the point $(15,36)$ is on the line of direct variation

therefore

the answer is

$(0,0)$

$(2.5,6)$

$(7.5,18)$

$(15,36)$

5 0

2 years ago

Read 2 more answers

The number of bees that visit a plant is 500 times the number of years the plant is alive, where t represents the number of year

user100 [1]

here we have given that number of bees visit a plant id 500 times the number of years that the plant is alive.

we know that t is the number of years that plant is alive.

we know that $500*t$ is expression representing the number of bees that will visit the plant in its life time.

5 0

2 years ago

Shana bought 6.2 pounds of pecans and paid $56.00. About how much per pound did the pecans cost?

bogdanovich [222]

Answer:

D. About $10

Step-by-step explanation:

To find unit price, you divide the total price by how much you're buying.

For example - $56 divided by 6.2 pounds of pecans is 9.032258064516129032258064516129‬ which can be rounded up to about $10

7 0

2 years ago

Read 2 more answers

A farmer has 300 ft of fencing with which to enclose a rectangular pen next to a barn. The barn itself will be used as one of th

Katarina [22]

Let x represent the length of the side of the pen that is parallel to the barn. Then the area (y) will be
.. y = x(300 -x)/2
This describes a downward opening parabola with zeros at x=0 and x=300. The vertex (maximum) will be found at the value of x that is halfway between those, x = 150.

For that value of x, the pen area is
.. y = 150(300 -150)/2 = 150^2/2 = 11,250 . . . . . square feet.

5 0

2 years ago

A train station has installed a system for determining whether bags contain explosives. It has a 95% chance of correctly identif

astraxan [27]

Answer:

a) P(identified as containing explosives)=P(actually contains explosives and identified as containing explosives)+P(actually not contains explosives and identified as containing explosives)

=(10/(4*106))*0.95+(1-10/(4*106))*0.005 =0.005002363

hence probability that it actually contains explosives given identified as containing explosives)

=(10/(4*106))*0.95/0.005002363=0.000475

let probability of correctly identifying a bag without explosives be a

hence a =0.99999763 ~ 99.999763%

No as even if that becomes 1 ; proportion of true explosives will always be less than half of total explosives detected,

3 0

2 years ago