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2 years ago

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Which statements are true about the graph of y ≤ 3x + 1 and y ≥ –x + 2? Check all that apply. 1.The slope of one boundary line i

s 2.
2.Both boundary lines are solid.
3.A solution to the system is (1, 3).
4.Both inequalities are shaded below the boundary lines.
5.The boundary lines intersect.

2 answers:

zloy xaker [14]2 years ago

7 0

Answer:

2.Both boundary lines are solid.

3.A solution to the system is (1, 3)

5.The boundary lines intersect.

Step-by-step explanation:

we have

$y\leq 3x+1$ ----> inequality A

The solution of the inequality A is the shaded area below the solid line $y=3x+1$

The slope of the solid line is 3

The point (1,3) is a solution of inequality A (lie in the shaded area of the solution set)

$y\geq -x+2$ ----> inequality B

The solution of the inequality B is the shaded area above the solid line $y=-x+2$

The slope of the solid line is -1

The point (1,3) is a solution of inequality B (lie in the shaded area of the solution set)

The solution of the system of inequalities is the shaded area between the two solids lines

see the attached figure

<u><em>Verify each statement</em></u>

1.The slope of one boundary line is 2

The statement is False

2.Both boundary lines are solid.

The statement is True

3.A solution to the system is (1, 3)

The statement is True

4.Both inequalities are shaded below the boundary lines

The statement is False

5.The boundary lines intersect.

The statement is True

The intersection point is (0.25,1.75)

see the attached figure

mars1129 [50]2 years ago

4 0

Answer:

2.Both boundary lines are solid.

3.A solution to the system is (1, 3)

5.The boundary lines intersect.

Step-by-step explanation:

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Given that PS¯¯¯¯ is the perpendicular bisector of QR¯¯¯¯¯, PQ=12.4, PR=12.4, and SR=7.6, identify QR.

Kazeer [188]

Answer:

15.2

Step-by-step explanation:

As it is given that PS is perpendicular bisector of QR and PQ = PR = 12.4

and SR = 7.6

So,

QR = QS + SR

= 2(SR)

= 2×7.6

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2 years ago

Read 2 more answers

In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three

RUDIKE [14]

Answer:

The greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

Step-by-step explanation:

We can think of each displacement as a vector, with a given magnitude and direction.

A vector can be written using its x and y coordinates like this

$\vec{t}=(x, y)$

So, for the displacements a and c their vector coordinates would be:

$\vec{a}=(2, 0)$

$\vec{c}=(-1, 0)$

As the b displacement has an angle of 30° toward the north from due east, we can find its x and y coordinates using the following formulas:

$x=(magnitude)*cos(angle)$

$y=(magnitude)*sin(angle)$

Note: the angle in the formula is the one formed with the east measured counterclockwise.

So, the x and y coordinates for the b displacement will be:

$\vec{b}=(2*cos(30), 2*sin(30))$

As the vector additon is commutative, the order won't affect the final position. Nevertheless, any change in the direction of any displacement will change the final position. So, in order to find the combination greatest distance we should calculate the following additions and find the one with the greatest magnitude:

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

Each resultant vector can be found adding each component. Afterwards, the magnitude can be found using the following formula:

$|\vec{R}|=\sqrt[ ]{(R_{x})^2 +{(R_{y})^2}}$

Now, let's calculate!

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$R_{1_x}} =2+2*cos(30)-1=2.73$

$R_{1_y}} =0+2*sin(30)+0=1$

$\vec{R_{1}}=(2.73,1)$

$|\vec{R_{1}}|=\sqrt[ ]{(2.73)^2 +{(1)^2}}=3.86$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$R_{2_x}} =2-2*cos(30)-1=0.73$

$R_{2_y}} =0-2*sin(30)+0=-1$

$\vec{R_{2}}=(-0.73,-1)$ }

$|\vec{R_{2}}|=\sqrt[ ]{(-0.73)^2 +{(-1)^2}}=1.03$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$R_{3_x}} =2+2*cos(30)+1=4.73$

$R_{3_y}} =0+2*sin(30)-0=1$

$\vec{R_{3}}=(4.73,1)$

$|\vec{R_{3}}|=\sqrt[ ]{(4.73)^2 +{(1)^2}}=6.69$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

$R_{4_x}} =2-2*cos(30)+1=1.26$

$R_{4_y}} =0-2*sin(30)-0=1$

$\vec{R_{4}}=(1.26,-1)$

$|\vec{R_{4}}|=\sqrt[ ]{(1.26)^2 +{(-1)^2}}=1.79$

So, after all the calculation, we can know for sure that the vector $\vec{R_{3}}$ has the biggest magnitude. Then, the greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

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2 years ago

A veterinarian needs to know an animal's weight in kilograms.If 20 pounds is about 9 kilograms and dog weighs 30 pounds,use a ra

Lesechka [4]

Answer:

Using the ratio table the dogs weight is:

30 pounds = 13.5 kilograms

Step-by-step explanation:

For this case we have the following conversion:

20 pounds = 9 kilograms

To use the table what we must do is find another relationship that allows us to find the weight in kilograms for 30 pounds.

For example, half the weight in pounds is half the weight in kilograms.

Therefore, the given conversion is:

10 pounds = 4.5 kilograms

So, for 30 pounds, we multiply this last ratio obtained by three on both sides:

30 pounds = 13.5 kilograms

Then, the table is:

Pounds 10 20 30

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2 years ago

A firm decides to give 20% extra free in their packets of soap powder. How much extra soap powder would be given away free with

zvonat [6]

Answer:

0.24 kg

Step-by-step explanation:

Remember that

$20\%=20/100=0.20$

To find out how much extra soap powder will be given away with the package, multiply the original weight by the percentage in decimal form.

so

$1.2(0.20)=0.24\ kg$

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2 years ago

When dealing with the number of occurrences of an event over a specified interval of time or space and when the occurrence or no

jeyben [28]

Answer:

POISSON DISTRIBUTION

Step-by-step explanation:

When dealing with the number of occurrences of an event over a specified interval of time or space, the poisson distribution is often useful.

Poisson distribution is applicable if:

The probability of the occurrence of the event is the same for any two intervals of equal length.

The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence in any other interval.

The probability that two or more events will occur in an interval approaches zero as the interval becomes smaller.

Therefore, the appropriate probability distribution is POISSON PROBABILITY DISTRIBUTION.

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2 years ago