Explain why you cannot evaluate √−25 using real numbers.

1 answer:

6 0

We can not evaluate it with real numbers, because there is no real number that multiplies itself twice and gives -25.

The answer is an imaginary number.

√-25 = 5i where i = √-1

So it can only be computed using imaginary numbers.

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your going to build a 3 foot by 7 garden.if you want to surround it by a 1 foot wide brick border , how many square feet will yo

stiv31 [10]

Answer:

Step-by-step explanation:

Multiply 3 and 7 by 2 and you get 6 and 14 6+14=20 which is the perimeter

Hope that helps!

5 0

3 years ago

What is the total area of the regions between the curves y = 6x2 − 18x and y = −6x from x = 1 to x=3?.

kvasek [131]

Using integration, it is found that the area between the two curves is of 22 square units.

<h3>What is the area between two curves?</h3>

The area between two curves y = f(x) and y = g(x), in the interval from x = a to x = b, is given by:

$A = \int_{x = a}^{x = b} |f(x) - g(x)| dx$

In this problem, we have that:

$f(x) = 6x^2 - 18x, g(x) = -6x, a = 1, b = 3$ .

Hence, the area is:

$A = \int_{1}^3 |6x^2 - 12x| dx$

$A = |x^3 - 6x^2|_{x = 1}^{x = 3}$

Applying the Fundamental Theorem of Calculus:

$A = |3^3 - 54 - 1^3 + 6|$

$A = 22$

The area between the two curves is of 22 square units.

More can be learned about the use of integration to find the area between the two curves at brainly.com/question/20733870

8 0

3 years ago

Can someone show me the steps to solving this <br> (Rationalizing Denominator)

Crazy boy [7]

Answer:

$\huge \purple {\frac{2 \sqrt[3]{6} + \sqrt[3]{18} }{ 6} }$

Step-by-step Explanation:

$\huge \frac{2 + \sqrt[3]{3} }{ \sqrt[3]{6} } \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )}{ \sqrt[3]{6} } \times \frac{ \sqrt[3]{6} \times \sqrt[3]{6} }{\sqrt[3]{6} \times \sqrt[3]{6}} \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )}{ \sqrt[3]{6} } \times \frac{ \sqrt[3]{6 ^{2} } }{\sqrt[3]{6^{2}} } \\ \\ = \huge \frac{(2 + \sqrt[3]{3} )\sqrt[3]{6} }{ \sqrt[3]{6} \times \sqrt[3]{6 ^{2} }} \\ \\ = \huge \frac{(2 \sqrt[3]{6} + \sqrt[3]{3} \sqrt[3]{6} )}{ \sqrt[3]{6 ^{3} }} \\ \\ = \huge \orange{\frac{2 \sqrt[3]{6} + \sqrt[3]{18} }{ 6} }$