Mathematics, 17.07.2019 07:00 tatia65

Which system of linear inequalities has the point (2, 1) in its solution set?

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You haven't provided the choices, therefore, I cannot provide an exact answer. However, I will help you with the concept.

For an order pair to be a solution to a system of equations, it has to satisfy BOTH equations. If it satisfies only one equation of the system or satisfy neither of the equations, the, it is not a solutions

Examples:
System 1:
x = y + 1
2x + 3y = 7
Let's check (2,1)
2 = 1 + 1 > equation 1 is satisfied
2(2) + 3(1) = 7 > equation 2 is satisfied
(2,1) is a solution to this system

System 2:
y = x + 3
y = x - 1
Let's check (2,1):
1 ≠ 2 + 3 > equation 1 isn't satisfied
1 = 2 - 1 > equation 2 is satisfied
(2,1) isn't a solution to this system

System 3:
2y = 9 - 3x
3x + 2y = 9
Let's ceck (2,1):
2(1) ≠ 9 - 3(2) > equation 1 isn't satisfied
3(2) + 2(1) ≠ 9 > equation 2 isn't satisfied
(2,1) isn't a solution to this system

Based on the above, all you have to do is substitute with (2,1) in the system you have and pick the one where both equations are satisfied

Hope this helps :)

Answer from: Quest

a. turk and mcalister

b. howard, mission, and market

c. yes. since streets end some do not cross yet are not parallel to others. example: mcalister and 5th

step-by-step explanation:

parallel lines are lines which do not intersect ever and look a lot like railroad tracks. turk, mcalister and golden gate are examples of these.

intersect are streets which cross each other and a car on one can reach the other. howard, mission, and market intersect with 6th.

skew lines are lines which neither intersect nor are parallel. these lines exist here because some streets end and never intersect another like mcalister and 5th.

A. name the streets that are parallel to golden gate. b. name the streets that intersect 6th street.

Answer from: Quest

d. 8

step-by-step explanation:

the given radical expression is:

$\sqrt{16}\times \sqrt{4}$

we can rewrite this as:

$(16)^{\frac{1}{2}\times (4)^{\frac{1}{2}$

we express as square numbers to get;

$(4^2)^{\frac{1}{2}\times (2^2)^{\frac{1}{2}$

the exponents will now cancel out to give.

$4\times2=8$

the correct answer is d

What is the value of the radical expression shown below

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