How to add subtract multiply and divide radical expressions

Mathematics

1 answer:

dem82 [27]3 years ago

7 0

Heres some info i learend

s but its not complete maybe i guess

Terms. We combine em by adding. The numbers or subtracting the numbers that are multiplied times the radicals.

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Identify the missing symbol in the expression. -0.5 ? -2/4

Alina [70]

The missing symbol is an equal sign

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3 years ago

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What is -0.8 as a fraction?

stira [4]

Answer:

$-\frac{4}{5}$

Step-by-step explanation:

There is not much that can be done to figure out how to write 0.8 as a fraction, except to literally use what the decimal portion of your number, the .8, means.

Since there are 1 digits in 8, the very last digit is the "10th" decimal place.

So we can just say that .8 is the same as 8/10.

The fraction is not reduced to lowest terms. We can reduce this fraction to lowest terms by dividing both the numerator and denominator by 2.

Why divide by 2? 2 is the Greatest Common Divisor (GCD) or Greatest Common Factor (GCF) of the numbers 8 and 10.

So, this fraction reduced to lowest terms is 4/5

So your final answer is: 0.8 can be written as the fraction 8/10 simplified to 4/5

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3 years ago

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Find the fifth roots of 243(cos 260° + i sin 260°)

Charra [1.4K]

<span>use De Moivre's Theorem:

⁵√[243(cos 260° + i sin 260°)] = [243(cos 260° + i sin 260°)]^(1/5)

= 243^(1/5) (cos (260 / 5)° + i sin (260 / 5)°)

= 3 (cos 52° + i sin 52°)

z1 = 3 (cos 52° + i sin 52°) ←← so that's the first root

there are 5 roots so the angle between each root is 360/5 = 72°

then the other four roots are:

z2 = 3 (cos (52 + 72)° + i sin (52+ 72)°) = 3 (cos 124° + i sin 124°)

z3 = 3 (cos (124 + 72)° + i sin (124 + 72)°) = 3 (cos 196° + i sin 196°)

z4 = 3 (cos (196 + 72)² + i sin (196 + 72)°) = 3 (cos 268° + i sin 268°)

z5 = 3 (cos (268 + 72)° + i sin (268 + 72)°) = 3 (cos 340° + i sin 340°) </span>

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3 years ago

I need help please

julsineya [31]

Answer:

-2x = 4 4 + -2x

Step-by-step explanation:

-4x - 2x 2x - 4x

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3 years ago

Question 4

Elanso [62]