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Pavlova-9 [17]
3 years ago
15

Cube root of b to the 27 power​

Mathematics
1 answer:
TiliK225 [7]3 years ago
7 0

\sqrt[3]{b^{27}}=\sqrt[3]{(b^9)^3}=b^9

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Answer:

The first four.

Step-by-step explanation:

There are 3 main postulates. SSS, SAS, and AAS. This simply refers to how we prove a triangle congruent. With SSS, all 3 sides must be congruent (either proven or given). AAS is when you have 2 angles congruent with a side next to one of the angles. NOT IN BETWEEN (there's an image as to what I'm talking about below). Finally, SAS. This is when you have a set of angles congruent with sides on each side congruent as well (look at the first four as an example of this.

Any more specific questions, feel free to ask!

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Show that the equation x^4/2021 − 2021x^2 − x − 3 = 0 has at least two real roots.
andreev551 [17]

The roots of an equation are simply the x-intercepts of the equation.

See below for the proof that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

The equation is given as: \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

There are several ways to show that an equation has real roots, one of these ways is by using graphs.

See attachment for the graph of \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

Next, we count the x-intercepts of the graph (i.e. the points where the equation crosses the x-axis)

From the attached graph, we can see that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} crosses the x-axis at approximately <em>-2000 and 2000 </em>between the domain -2500 and 2500

This means that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

Read more about roots of an equation at:

brainly.com/question/12912962

6 0
3 years ago
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