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2 answers:
Answer: (A)
Explanation: For there to be a solution at any one point, both inequalities must share a common point. This point will then satisfy both inequations, becoming a solution to both inequality.
Let's pick (A).
For there to be at least one solution, the first inequality must have a solution of
. Since the left side has clearly no points common to both graphs. Hence, no solution can exist that will satisfy both.
All of the other graphs have at least one point in common.
(B) has
(C) has
(D) has
Explanation: For there to be a solution at any one point, both inequalities must share a common point. This point will then satisfy both inequations, becoming a solution to both inequality.
Let's pick (A).
For there to be at least one solution, the first inequality must have a solution of
All of the other graphs have at least one point in common.
(B) has
(C) has
(D) has
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Answer:
The total price of pounds of apples is <u>$6.48</u>.
Step-by-step explanation:
Given:
Last week,the price of apples at a grocery store was 1.60 per pound.This week,apples at the same grocery store are on sale for a 10% discount.
Now, to find the total price of 4 1/2 pounds of apples this week at the grocery store.
The price of apples at a grocery store last week was = $1.60 per pound.
This week apples at the same grocery store is on sale of 10% discount.
Thus, the price of apples at a grocery store this week is:
So, the price of apples per pound this week is $1.44.
Now, to get the total price of pounds of apples by using unitary method:
If 1 pound of apples cost =
Then, pound of apples cost =
Therefore, the total price of pounds of apples is $6.48.
Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)