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3 years ago

(Past Due) Need Help

Mathematics

1 answer:

Charra [1.4K]3 years ago

8 0

I would say its the second one.

the student shouldve distributed the 2^3x+3 into 2^3x+9

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Find the standard form of the equation of the ellipse with the given characteristics. Vertices: (4, 0), (4, 14); endpoints of th

saul85 [17]

Answer:

The standard form of the ellipse is $\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$ .

Step-by-step explanation:

The major axis of the ellipse is located in the y axis, whereas the minor axis is in the x axis. The center of the ellipse is the midpoint of the line segment between vertices, this is:

$(h, k) =\frac{1}{2}\cdot V_{1} (x,y) + \frac{1}{2}\cdot V_{2} (x,y)$ (1)

If we know that $V_{1} (x,y) = (4,0)$ and $V_{2}(x,y) = (4, 14)$ , then the coordinates of the center are, respectively:

$(h,k) = \frac{1}{2}\cdot (4, 0) + \frac{1}{2}\cdot (4,14)$

$(h,k) = (2,0) + (2, 7)$

$(h, k) = (4, 7)$

The length of each semiaxis is, respectively:

$a = \sqrt{(1 - 4)^{2}+(7-7)^{2}}$

$a = 3$

$b = \sqrt{(4-4)^{2}+(0-7)^{2}}$

$b = 7$

The standard equation of the ellipse is described by the following formula:

$\frac{(x-h)^{2}}{a^{2}}+ \frac{(y-k)^{2}}{b^{2}} = 1$

Where:

$h$ , $k$ - Coordinates of the center of the ellipse.

$a$ , $b$ - Length of the orthogonal semiaxes.

If we know that $h = 4$ , $k = 7$ , $a = 3$ and $b = 7$ , then the standard form of the ellipse is:

$\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$

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3 years ago

I'd appreciate it if anyone could help ASAP. I'm giving Brainliest!

n200080 [17]

Answer:

(1, 0.25)

Step-by-step explanation:

To find when f(x)=g(x), then look for a value that is the same for both functions in the table. 0.25 occurs twice for the same x value. This is when they are equal. The solution is (1, 0.25).

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The answer would be around $20

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