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2 years ago

Explain relationship between how radical expressions and rational (fractional) exponents

Mathematics

1 answer:

RideAnS [48]2 years ago

3 0

Answer:

Any radical in the form can be written using a fractional exponent in the form . The relationship between and works for rational exponents that have a numerator of 1 as well. For example, the radical can also be written as , since any number remains the same value if it is raised to the first power.

Step-by-step explanation:

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Bowl A contains 6 apples and 3 mangoes this can be written as, Bowl B contain 8 apples and 2 mangoes this can be written as. Wha

AlexFokin [52]

Answer:

2 apples and 1 mango

Step-by-step explanation:

Bowl A can be written as 6a + 3m

Bowl B can be written as 8a + 2m

The difference between the bown is 2 apples and 1 mango

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3 years ago

Martin charges $10 for every 5 bags of leaves he rakes. Last weekend, he raked 2 bags of leaves. How much money did he earn?

Gnoma [55]

He earned 4 dollars.

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3 years ago

Read 2 more answers

Help me figure this out

Semenov [28]

Answer:

Each number matches up below. The left side is x, and the right is y.

1 ⇒ 192

2 ⇒ 384

4 ⇒ 768

6 ⇒ 1152

10 ⇒ 1920

Step-by-step explanation:

We know in 1 serving, there is 192 calories. Following this, to find the calories for 2 servings, we multiply 192 by 2.

192*2 = 384

When we know the calories but not the serving, you divide the calories by 192

768/192 = 4

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2 years ago

Dude sombody pleasee help me nobody is taking it seriously:(

Mnenie [13.5K]

Answer:

(9, 5) or (0, 14)

Step-by-step explanation:

4+2+3+2=9

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3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

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3 years ago