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3 years ago

Can yall help me????

Mathematics

1 answer:

Kryger [21]3 years ago

3 0

Hi There!

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Remember to follow PEMDAS.

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Problem #1:

= (25 - 6.25) + (5 · 3)²

= (18.75) + (15)²

= 18.75 + 225

= 243.75

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Problem #2:

= (8 + 3)² + 10 · (-4)

= (11²) + 10 · (-4)

= 121 + 10 · (-4)

= 121 + (-40)

= 81

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Hope This Helps :)

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In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

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3 years ago

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seropon [69]

Area of the shaded region = Area of a square - Area of a circle

Side of a square = 2r
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Area of a circle = πr²

Area of the shaded region = 4r² - πr² = r²(4-π)

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