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Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

miss Akunina [59]

Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

Equation:

$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

$y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

7 0

3 years ago

Read 2 more answers

Tara and Jody's bedrooms are shaped like rectangles. Tara's bedroom is 9 ft long and 8 ft wide. Jody's bedroom is 7ft long and 1

horrorfan [7]

Tara's Bedroom: A = lw
A = (9)(8)
A = 72 ft²

Jody's Bedroom: A = lw
A = (7)(10)
A = 70 ft²

Tara's bedroom has the greater area than Jody's bedroom.

5 0

3 years ago

Please help with this

11Alexandr11 [23.1K]

The answer is

4x^4-4x^3-16x^2+16x

4 0

3 years ago

Ben and Sam are diving into a lake. At 14 below the surface, Ben spots Sam 9 ft directly below him. Find Sams depth SHOW WORK AN

natka813 [3]

Sam is at 23 feet below the surface

Explanation
I did 14+9 and got 23

7 0

2 years ago

Read 2 more answers

Evaluate the definite integral using the graph of f(x)<br> (Image included)

Tanya [424]

a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is

$\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}$

b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so

$\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}$

c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so

$\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20$

Then by linearity, we have

$\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}$

8 0

2 years ago