If altitude BD for ABC falls outside the triangle, what kind of triangle is ABC?
2 answers:
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Answer:
(8 - 12 )
Step-by-step explanation:
Given
x -
y = 8
x +
y = - 5
Multiply both equations through by 4 to clear the fractions
x - 2y = 32 → (1)
2x + 3y = - 20 → (2)
Multiplying (1) by - 2 and adding to (2) will eliminate the x- term
- 2x + 4y = - 64 → (3)
Add (2) and (3) term by term to eliminate x
0 + 7y = - 84
7y = - 84 ( divide both sides by 7 )
y = - 12
Substitute y = - 12 into either of the 2 equations and solve for x
Substituting into (1)
x - 2(- 12) = 32
x + 24 = 32 ( subtract 24 from both sides )
x = 8
solution is (8, - 12 )
Answer:
C. x^2 + 2x - 1 = 3
Step-by-step explanation:
The standard form of a quadratic equation is
ax^2 + bx + c = 0
We need to use the quadratic formula and the given expression to find the values of a, b, and c.
The quadratic formula is
Where the formula has -b, the problem has -2, so b = 2.
Now we have
ax^2 + 2x + c = 0
In the denominator, where the formula has 2a, the problem has 2(1), so a = 1.
Now we have
x^2 + 2x + c = 0
Inside the root, the quadratic formula has -4ac. the problem shows -4(1)(-4). Since we already know that a = 1, then c = -4.
Now we have
x^2 + 2x - 4 = 0
Let's look at choice A.
x^2 + 1 = 2x - 3
Subtract 2x from both sides. Add 3 to both sides.
x^2 - 2x + 4 = 0 <em>This is not it!</em>
Let's look at choice B.
x^2 - 2x - 1 = 3
Subtract 3 from both sides.
x^2 - 2x - 4 = 0 <em>This is not it!</em>
Let's look at choice C.
x^2 + 2x - 1 = 3
Subtract 3 from both sides.
x^2 + 2x - 4 = 0 <em>This is it!</em>
Answer: C. x^2 + 2x - 1 = 3