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2 years ago

Todos los enteros son racionales

Mathematics

1 answer:

Stels [109]2 years ago

6 0

Answer:

si si si si si si los números racionales también son fracciones

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The lcm of 5 and 8 is??

zysi [14]

The least common multiple of 5 and 8 is 40. This is because when you go by multiples of 5 and 8 the multiple that is the “smallest number” (LCM) of both numbers is 40. :)

7 0

2 years ago

The radius of a sphere is increasing at a rate of 3 mm/s. how fast is the volume increasing when the diameter is 100 mm?

Katena32 [7]

33 per seconds
1s/3mm x/100mm x=33.3 s

4 0

2 years ago

Mina says triangle A with angles

kotykmax [81]

Answer:

Yes, Mina is correct

Step-by-step explanation:

Let the triangles be A and B

Given

Triangle A:

$Angles = 34, 57$

Triangle B:

$Angles = 89, 57$

Required

Is Mina's claim correct?

First, we calculate the third angle in both triangles.

For A:

$Third\ Angle = 180 - 34 - 57$

$Third\ Angle = 89$

For B:

$Third\ Angle = 180- 89 - 57$

$Third\ Angle = 34$

For triangle A, the angles are: 34, 57 and 89

For triangle B, the angles are: 34, 57 and 89

<em>Since both triangles have the same angles, then by the postulate of AAA (Angle-Angle-Angle), the triangles are similar.</em>

3 0

2 years ago

The product of a number and -3 is less than 27

Vaselesa [24]

Answer:

-8

Step-by-step explanation:

-8×-3 =24 thats yeaaaa

3 0

2 years ago

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

3 0

2 years ago