At a parking garage you can park underground or above ground. The lowest part of the underground parking is 40 feet below ground
1 answer:
You might be interested in
We know is a semi-circle, so let us use the equation for the area of a circle instead, and then, half it
so, the diameter is 1100, meaning the radius is half that, or 550
so...
so, the diameter is 1100, meaning the radius is half that, or 550
so...
The first problem:
y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴
The correct options are 2 , 7
Solution of the system of equations is
and
==================================
The second problem:
The general equation of the hyperbole is
For the given equation:
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36 ⇒⇒⇒ b = √36 = 6
∴ the slope of the <span><span>asymptotes = </span>b/a and -b/a
</span><span>
</span><span>b/a = 6/15 = 2/5
</span><span>
</span><span>-b/a = -6/15 = -2/5
</span><span>
</span><span>∴ m = 2/5 and m = -2/5
</span>
y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴
The correct options are 2 , 7
Solution of the system of equations is
and
==================================
The second problem:
The general equation of the hyperbole is
For the given equation:
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36 ⇒⇒⇒ b = √36 = 6
∴ the slope of the <span><span>asymptotes = </span>b/a and -b/a
</span><span>
</span><span>b/a = 6/15 = 2/5
</span><span>
</span><span>-b/a = -6/15 = -2/5
</span><span>
</span><span>∴ m = 2/5 and m = -2/5
</span>