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3 years ago

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Mathematics

1 answer:

Finger [1]3 years ago

4 0

Answer:

The answer to this question can be defined as follows:

Step-by-step explanation:

In the given question some information is missing that is values. If we assume the value which can be described as follows:

$5\geq 6$

In the above equation it will give 6 value because 5 is not greater then equal to 6.

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Combine and simplify the following radical expression ASAP ASAP ASAP ASAP

Montano1993 [528]

Answer:

$\sqrt[3]{3}$

Step-by-step explanation:

Our expression is: $\frac{1}{3} \sqrt[3]{81}$ .

Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or $3^3*3$ . Put this in for 81:

$\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}$

We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:

$\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}$

Now, let's multiply this by 1/3, as shown in the original problem:

$\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}$

Thus, the answer is $\sqrt[3]{3}$ .

~ an aesthetics lover

6 0

3 years ago

Round 41.805 to the nearest whole number

aleksandr82 [10.1K]

The answer to your question is 42. The rule is 5 or more go up 1 . If the number is 4 les stay where it is. hope this help.

5 0

3 years ago

Read 2 more answers

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of

notsponge [240]

Answer:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, $(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)$

So, The coordinates of the triangle A'B'C' are:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

6 0

3 years ago

Write a question that matches this equation: 45-n=15

Alja [10]

Answer:

ayla had 45 cupcakes for the class jayla gave a number out to the cladd Jayla took 15 cupcakes back home what is the does n equal to

6 0

3 years ago

What is the measure of Z6?

DedPeter [7]

Yeebcxccfsfhgwsrthjrss

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3 years ago