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artcher [175]
3 years ago
11

Can Fractions and ratios cannot have zero in the denominator?

Mathematics
1 answer:
labwork [276]3 years ago
4 0
True , is the answer ! :)
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The anwser is going to be six 6
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10×2 tens =? tens = ?
Dmitry_Shevchenko [17]
20 tens= 2 tens = 20 ones
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LaTeX: \sqrt[]{36} 36 2) LaTeX: \sqrt[3]{-8} − 8 3 3) LaTeX: -\sqrt[]{100} − 100 4) LaTeX: \sqrt[3]{27} I'm sorry that's the bes
Mamont248 [21]

Given:

Consider the expression are

1) \sqrt{36}

2) \sqrt[3]{-8}

3) -\sqrt{100}

4) \sqrt[3]{27}

To find:

The simplified form of each expression.

Solution:

1. We have,

\sqrt{36}=\sqrt{6^2}

\sqrt{36}=6

Therefore, the value of this expression is 6.

2. We have,

\sqrt[3]{-8}=(-8)^{\frac{1}{3}}

\sqrt[3]{-8}=((-2)^3)^{\frac{1}{3}}

\sqrt[3]{-8}=(-2)^{\frac{3}{3}}

\sqrt[3]{-8}=-2

Therefore, the value of this expression is -2.

3. We have,

-\sqrt{100}=-\sqrt{10^2}

-\sqrt{100}=-10

Therefore, the value of this expression is -10.

4. We have,

\sqrt[3]{27}=(27)^{\frac{1}{3}}

\sqrt[3]{27}=(3^3)^{\frac{1}{3}}

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

\sqrt[3]{27}=3

Therefore, the value of this expression is 3.

3 0
3 years ago
In a triangle the measurements of the angles are x°, ( x + 6)°, (2x + 10)°, and the sum of the three angles of a triangle is 180
34kurt

Answer:

largest angle = 92°

Step-by-step explanation:

Since the measurement of the angles are;

x°, ( x + 6)°, (2x + 10)°

And we know that sum of angles in a triangle is 180. Thus;

x + (x + 6) + (2x + 10) = 180

x + x + 6 + 2x + 10 = 180

4x + 16 = 180

4x = 180 - 16

4x = 164

x = 164/4

x = 41°

The other two angles will be;

(41 + 6)° and (2(41) + 10)°

Which gives;

47° and 92°

So,largest angle = 92°

7 0
3 years ago
Randy is starting a beehive so that he can have fresh
klemol [59]

Answer:

1,280

Step-by-step explanation:

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2 years ago
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