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3 years ago

PROVING THEOREMS OF SQUAD can someone help me pleasee ASAP

Mathematics

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

40 units

Step-by-step explanation:

For a square, all the sides are equal and the interior angles are equal and all equal to 90. Hence;

m<BOJ = 90 degrees

m<BOJ = 4x - 6

Equating both to get x;

4x - 6 = 90

4x = 90+6

4x = 96

x = 96/4

x = 24

Since all the sides are equal, hence BO = JO = 2x-8

JO = 2x - 8

Substitute x = 24 into JO

JO = 2(24) - 8

JO = 48 - 8

JO = 40

Hence the measure of JO is 40 units

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