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3 years ago

Plz :<

Mathematics

1 answer:

maria [59]3 years ago

6 0

Answer:

the answer must be A

Step-by-step explanation:

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Find the value of X in the triangle shown below

QveST [7]

Answer:

102

Step-by-step explanation

39'+39'=78'

180-78=102

7 0

2 years ago

Difference in the following polynomials (6x^3 - 2x^2 + 4) -(2x^3 + 4x^2 - 5)

Colt1911 [192]

$\bold{[ \ Answer \ ]}$

$\boxed{\bold{4x^3-6x^2+9}}$

$\bold{[ \ Explanation \ ]}$

$\bold{Find \ Difference: \ \left(6x^3\:-\:2x^2\:+\:4\right)-\left(2x^3\:+\:4x^2\:-\:5\right)}$

$\bold{-----------------------}$

$\bold{Remove \ Parenthesis}$

$\bold{6x^3-2x^2+4-\left(2x^3+4x^2-5\right)}$

$\bold{Remove \ Parenthesis \ / \ Simplify}$

$\bold{-\left(2x^3+4x^2-5\right): \ -2x^3-4x^2+5}$

$\bold{Rewrite}$

$\bold{6x^3-2x^2+4-2x^3-4x^2+5}$

$\bold{[Simplify \ 6x^3-2x^2+4-2x^3-4x^2+5] \ 4x^3-6x^2+9}$

$\bold{4x^3-6x^2+9}$

$\boxed{\bold{[] \ Eclipsed \ []}}$

6 0

3 years ago

Find the inverse or opposite of (2n-8)

asambeis [7]

the additive inverse of -8 is 8

to find the additive inverse you just have to change the sign of the number

5 0

3 years ago

If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

Step 2: Differentiate

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$