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1 year ago

Simplify the expression 5^3 x 5^-5 A. 5^2 B. 1/5 C. 1/5^2 D. -52

Mathematics

1 answer:

jeka941 year ago

5 0

${5}^{3} \: \times \: {5}^{ - 5}$

We first compute the product.

${5}^{ - 2}$

We can transform the product into positive if we use this formula:

$\boxed{ {a}^{ - n} \: = \: \frac{1}{ {a}^{n} } }$

<h3>We apply it:</h3>

${5}^{ - 2} \: = \: \boxed{ \bold{\frac{1}{ {5}^{2} } }}$

<h2>Answer: </h2>

$\text{Option C.} \: \boxed{ \bold{\frac{1}{ {5}^{2} } }}$

<h3>MissSpanish</h3>

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3 years ago

Write (8a-^3) -2/3 in simplest form

Anna71 [15]

$(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Solution:

Given that,

$(8a^{-3})^{\frac{-2}{3}$

We have to write in simplest form

Use the following law of exponent

$(a^m)^n = a^{mn}$

Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Thus the given expression is simplified

8 0

3 years ago

Solve -5x2 - 20x + 35 = 30 by completing the square.

Semenov [28]

Answer:

x=−2−√5 or x=−2+√5

Step-by-step explanation:

-5x2 - 20x + 35 = 30

Answer = x=−2−√5 or x=−2+√5

5 0

2 years ago

1. Derive the half-angle formulas from the double

lilavasa [31]

1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The linear function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following trigonometric formulae are used to derive the half-angle formulas:

sin² θ / 2 + cos² θ / 2 = 1 (1)

cos θ = cos² (θ / 2) - sin² (θ / 2) (2)

First, we derive the formula for the sine of a half angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a half angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a half angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in rectangular form to polar form are obtained by trigonometric functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in polar form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in rectangular form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the linear function y = 5 · x - 8, we proceed to use the following substitution formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The linear function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

#SPJ1

4 0

2 years ago