3 years ago

How do we find the second and the third derivative of this equation? Please help!

Mathematics

1 answer:

Gelneren [198K]3 years ago

5 0

Answer:

$\frac{7}{2\sqrt{2} }$

Step-by-step explanation:

substitute 2 in for x. the square root of 2 is already in radical form

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2 years ago

If the measure of angle 1 is (3 x minus 4) degrees and the measure of angle 2 is (4 x + 10) degrees, what is the measure of angl

arlik [135]

Answer:

58 degrees

Step-by-step explanation:

If angle 1 + angle 2 = 90 degrees, then:

(3x - 4) + (4x+10) = 90

Combine like terms:

(3x+4x) + (10-4) = 90

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2 years ago

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2 years ago

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Fill in the missing portions of the function to rewrite g(x) = 3a^2 − 42a + 135 to reveal the zeros of the function. What are th

mariarad [96]

Answer:

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

Step-by-step explanation:

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:

$g(x) = 3x^{2} - 42a + 135$

$a = 3, b = -42, c = 135$

$\bigtriangleup = (-42)^{2} - 4*3*135 = 144$

$x_{1} = \frac{-(-42) + \sqrt{144}}{2*3} = 9$

$x_{2} = \frac{-(42) - \sqrt{144}}{2*3} = 5$

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

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3 years ago

Cos 60 degrees = 1/2 and sin 60 degrees = square root 3/2 O A. True O B. False

garik1379 [7]

Step-by-step explanation:

sin(60)=

$\sqrt{3 |2| }$

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2 years ago

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