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yKpoI14uk [10]
4 years ago
5

PLEASE SOLVE THE QUESTIONS (WILL MARK BRANLIEST)

Mathematics
1 answer:
mel-nik [20]4 years ago
7 0

Answer:

a) 4⁻³ = 1/64 = 0.015625

b)13⁻² = 1/169 = 0.0059171598

c)(-3)⁻² = 1/-3² = 0.1111111111

Step-by-step explanation:

To solve the question above, when we have an integer ( positive or negative) that is raised to a negative power, this means the reciprocal of that integer raised to the positive power

Example:

a⁻ⁿ = 1/aⁿ

a) 4⁻³ = 1/4³

= 1/(4 × 4 × 4)

= 1/64

= 0.015625

b) 13⁻² = 1/13²

= 1/(13 × 13)

= 1/169

= 0.0059171598

c)(-3)⁻² = 1/-3²

= 1/(-3 × -3)

= 1/9

= 0.1111111111

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3 years ago
Find the indicated value. Use either half angle formulas or sum and difference formulas.
MrMuchimi
By the angle difference formula for sines, we have:

\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ = \frac{sqrt{6}}{4} - \frac{sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.

By the angle sum formula for tangents, we have:

\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1+\frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}}.

Rationalizing the denominator gives \frac{12+6\sqrt{3}}{6} = 2+\sqrt{3} as the final answer.
8 0
4 years ago
Solve: x + 3 = -x + 7 (5 points)
makvit [3.9K]

Answer:

x = 2

Step-by-step explanation:

x + 3 = -x + 7

Add x to each side

x+x + 3 = -x+x + 7

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7 0
4 years ago
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Quadratic Formula to solve the equation x^2- 4x = - 7
mylen [45]

roots of the equation x^2- 4x = - 7 are  2+3i and 2-3i

What are the natures of root of quadratic equation?

Case I: 4ac > b^2

The roots of the quadratic equation ax^2 + bx + c = 0are real and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive.

Situation II:  b^2-4ac =0

The roots and of the quadratic equationax^2 + bx + c = 0  are real and equal when a, b, and c are real numbers, a 0, and the discriminant is zero.

Case III: b^2-4ac<0

The quadratic equation ax^2 + bx + c = 0 has roots when a, b, and c are real numbers, a 0, and the discriminant is negative, but these roots are not equal and are not real. We refer to the roots in this instance as fictitious.

Case IV: Perfect square and b^2 - 4ac > 0

The roots of the quadratic equation ax^2 + bx + c = 0  are real, rational, and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive and perfect square.

Quadratic Formula x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

x^2- 4x = - 7

x^2-4x+7=0

Here a = 1 , b = -4 , c = 7

using quadratic formula

x={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(7)}}}{2(1)}}

x={\frac {4\pm {\sqrt {-12}}}{2}}

x={\ {2\pm {\ 3i}}

Learn more about quadratic equation from the link below

brainly.com/question/2279540

#SPJ1

8 0
1 year ago
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