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3 years ago

Will give brainliest PLEASE HELP!!! :D

Mathematics

2 answers:

Thepotemich [5.8K]3 years ago

8 0

I think A would be the answer

marshall27 [118]3 years ago

5 0

Answer:

Step-by-step explanation:

hope this helps

;)

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A circle centered at the origin has a radius of 10 units. The terminal side of an angle, Theta, intercepts the circle in Quadran

Tanzania [10]

Since the circle is centered at the origin, we can use the equations
x² + y² = r²
x = r cos θ
y = r sin θ

We know y = 8 and r = 10. Solving for x
x² + 8² = 10²
x = 6

From this equation:
x = r cos θ
6 = 10 cos θ
cos θ = 6/10 = 0.6

7 0

3 years ago

Reduce fraction: a^3+a^2b/5a times 25/3b+3a

Margaret [11]

ANSWER

$\frac{5a}{3}$

EXPLANATION

The given fractions are:

$\frac{{a}^{3} + {a}^{2} b}{5a} \times \frac{25}{3b + 3a}$

We factor to obtain:

$\frac{{a}^{2}(a + b)}{5a} \times \frac{25}{3(a + b)}$

We cancel the common factors to get:

$\frac{{a}(1)}{1} \times \frac{5}{3(1)}$

We multiply the numerators and also multiply the denominators to get:

$\frac{5a}{3}$

Therefore the two fractions simplifies to $\frac{5a}{3}$

3 0

3 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$