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faust18 [17]
3 years ago
13

−x − y = 1

Mathematics
2 answers:
Gemiola [76]3 years ago
5 0
<span>The system of linear equations

−x − y = 1
y = x + 3

can be solved in various ways.  This problem hints that you should graph both lines and find the coordinates of their point of intersection.

But we can also solve this system using substitution.  Note that y = x + 3.  We can subst. x + 3 for y in the 1st equation, as follows:

-x - (x+3) = 1.  This simplifies to -x -x - 3 = 1, or -2x = 4, or x = -2.

Since y = x + 3, subbing -2 for x produces y:  y = -2 + 3 = 1

Thus, the solution is (-2, 1). </span>
OLEGan [10]3 years ago
5 0
<span>−x − y = 1  so y = -x- 1
y = x + 3 
-x - 1 = x + 3
2x = -4
  x = -2

y = -2 + 3 = 1

(-2, 1)

answer
</span><span>(−2, 1)</span>
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The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

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