What is the vertex?? Please answer QUICK!

Mathematics

1 answer:

astra-53 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

To obtain vertex form from the given equation use the method of completing the square

given 8y + x² = - y² - 181 + 26x

rearrange having the x terms and y terms together

add y² to both sides

8y + x² + y² = - 181 + 26x ( subtract 26x from both sides )

x² - 26x + y² + 8y = - 181

add (half the coefficient of the x/y terms )² to both sides

x² + 2(- 13)x + 169 +y² + 2(4)y + 16 = - 181 + 169 + 16

completing the square on both the x and y terms

(x - 13)² + (y + 4)² = 4 → D

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Which is bigger, <br> a) 70% of 300 <br> b) ¼ of 600?<br> ∞Δ∞

Artist 52 [7]

Since 300 is 100%, you treat it as such.

300/(100/70) = 210 (rounded)

You do a similar thing for 600

1/4 of 100 is 25
100/25 = 4
600 / 4 = 150

70% of 300 is more.

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3 years ago

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stiv31 [10]

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2 years ago

A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one comp

Morgarella [4.7K]

Answer:

$y = 11 - 10cos(0.1\pi t)$

Step-by-step explanation:

Suppose at t = 0 the person is 1m above the ground and going up

Knowing that the wheel completes 1 revolution every 20s and 1 revolution = 2π rad in angle, we can calculate the angular speed

2π / 20 = 0.1π rad/s

The height above ground would be the sum of the vertical distance from the ground to the bottom of the wheel and the vertical distance from the bottom of the wheel to the person, which is the wheel radius subtracted by the vertical distance of the person to the center of the wheel.

$y = h_g + d_b = h_g + R - d_c$ (1)

where $h_g = 1$ is vertical distance from the ground to the bottom of the wheel, $d_b$ is the vertical distance from the bottom of the wheel to the person, R = 10 is the wheel radius, $d_c$ is the vertical distance of the person to the center of the wheel.

So solve for $d_c$ in term of t, we just need to find the cosine of angle θ it has swept after time t and multiply it with R

$d_c = Rcos(\theta(t)) = Rcos(\omega t) = 10cos(0.1\pi t)$

Note that $d_c$ is negative when angle θ gets between π/2 (90 degrees) and 3π/2 (270 degrees) but that is expected since it would mean adding the vertical distance to the wheel radius.

Therefore, if we plug this into equation (1) then

$y = h_g + R - d_c = 1 + 10 - 10cos(0.1\pi t) = 11 - 10cos(0.1\pi t)$