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3 years ago

What is 1.452 rounded to the nearest hundredth

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

8 0

Answer:

1.450

Step-by-step explanation:

what grade ar4e you in I'm guessing 4th or 5th

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Conjugate/Rational Number?

hram777 [196]

Answer:

1) $\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) $\dfrac{2}{\sqrt{5} }$

$\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

$\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} }$

$-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

$-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} }$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } \times \dfrac{ \sqrt{10} }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50} }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2} }{10} = \dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} }$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6} }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} } = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

6) $\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} }$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} =\dfrac{\sqrt{21} + \sqrt{35}}{8}$

8 0

3 years ago

A 12 foot ladder is leaning against a building . The ladder makes a 45 degree angle with the building . How far up the building

Firlakuza [10]

May this method work

Mark as brainliest give 5 star

8 0

3 years ago

Read 2 more answers

Write the following numbers in order from least to greatest.

creativ13 [48]

Answer:

52.6,56.6,594.3

Step-by-step explanation:

you multiply all o them

4 0

3 years ago

Please help being timed!!! Given: △ABC AB=BC=AC Area of △ABC= 25 3 4 cm2 Find: The perimeter

nirvana33 [79]

Answer:

15 cm

Step-by-step explanation:

The area of an equilateral triangle is:

$\frac{\sqrt{3} }{4}a^2$ , where a is the length of one side.

We can find the value of a (the length) by:

$\frac{25\sqrt{3} }{4} = \frac{\sqrt{3} }{4}a^2$

$a^2 = 25$

a = 5, so the side length of this triangle is 5 cm.

The perimeter of an equilateral triangle will be 3 times the length.

$3\cdot5=15$

Hope this helped!

7 0

3 years ago

Read 2 more answers

Helpppppppppppppppppppppp

Ad libitum [116K]

Answer:

8.04 units

Step-by-step explanation:

By Pythagoras Theorem:

${x}^{2} = \bigg( \frac{25.2}{2} \bigg)^{2} - {(9.7)}^{2} \\ \\ {x}^{2} = ( 12.6)^{2} - {(9.7)}^{2} \\ \\ {x}^{2} = 64.67 \\ \\ x = \sqrt{64.67} \\ \\ x = 8.04176598 \\ \\ x \approx \: 8.04$

4 0

2 years ago