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3 years ago

The minimum turning point

Mathematics

1 answer:

ycow [4]3 years ago

3 0

Answer:

y = x² - 6x + 7

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the turning point and a is a multiplier

Here (h, k) = (3, - 2 ) and letting a = 1

y = (x - 3)² - 2 ← expand using FOIL

= x² - 6x + 9 - 2 ← collect like terms

= x² - 6x + 7

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The distance between points (X7, Yı) and (4, 8) is the square root of (x7 - 8)2 +

serious [3.7K]

Answer:

B. False

Step-by-step explanation:

Given

$Points: (x_7 - y_i),(4,8)$

Required

Is the distance between them $\sqrt{(x_7 - 8)^2 + (y_1 - 4)^2}$

Given that there are two points, the distance between them is calculated using

Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

In this case;

$(x_1,y_1)= (x_7 - y_i)\\(x_2,y_2) =(4,8)$

Substitute these in the given formula

Distance = $\sqrt{(4- x_7)^2 + (8 - y_1)^2}$

Comparing this with the given formula, both are not the same;

Hence, the formula is false and the correct formula is of the form $\sqrt{(4- x_7)^2 + (8 - y_1)^2}$

6 0

3 years ago

The velocity of a particle moving back and forth on a line is vequalsStartFraction ds Over dt EndFraction equals6 sine (2 t )Sta

max2010maxim [7]

Answer:

s = 6 m

Step-by-step explanation:

The value of the velocity v is given as:

$v = \frac{ds}{dt} = 6 sin(2t)$ m/s

To find s, we have to integrate and apply the initial values of s = o when t = 0:

$\frac{ds}{dt} = 6 sin(2t)\\\\\int\limits^s_0 {ds} = \int\limits^t_0 {6sin(2t)} \, dt\\\\s|^s_o = -3cos(2t)|^t_o\\\\s - 0 = -3cos(2t) -(-3cos(0))\\\\s = -3cos(2t) + 3(1)\\\\s = -3cos(2t) + 3$

When t = π/2, s will be:

s = -3cos(2 * π/2) + 3

s = -3cos(π) + 3

s = -3(-1) + 3

s = 3 + 3

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4 years ago

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Answer:

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Step-by-step explanation:

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