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svp [43]
3 years ago
12

What is the solution to the inequality? X-7>-6

Mathematics
1 answer:
Likurg_2 [28]3 years ago
4 0
The equation is saying x minus 7 is greater than -6, so treat the inequality symbol just how you would with a normal =. So x-7=-6, you need to find x, so to find x you need to keep x alone. So you get rid of the -7, so to do that since it is a negative you add 7 to -6, so that would mean that x=1. Now you put the inequality symbol back. So now it's x>1. This is the solution, x is greater than 1, or x>1. Hope this helps
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Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
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
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d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ 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 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












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1 5/16 tons as a decimal
liubo4ka [24]
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6 0
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PQ is rotated 180 degrees clockwise about P. What are the coordinates of P' and Q'?
Salsk061 [2.6K]

The co ordinates of P' is (-7,-2) and Q' is (-16, -8)

<u><em>Explanation</em></u>

PQ is rotated 180 degrees clockwise about P. It means <u>P and P' are the same points</u>.

According to the graph, the coordinates of P is (-7, -2) and  Q is (2, 4)

When PQ is rotated 180 degrees clockwise about P, then <u>P or P' will be the mid-point of Q and Q' </u>

Suppose, the co ordinate of Q' is (x, y)

Now according to the mid-point formula, the coordinate of P or P' will be: (\frac{x+2}{2},\frac{y+4}{2}) , which is actually at (-7, -2)

Thus.....

\frac{x+2}{2}=-7\\ \\ x+2=-14\\ \\ x= -16\\ \\ and\\ \\ \frac{y+4}{2}= -2\\ \\ y+4= -4\\ \\ y= -8

So, the co ordinates of P' is (-7,-2) and Q' is (-16, -8)

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