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2 years ago

Rod made a scale drawing of a petting zoo. He used the

Mathematics

1 answer:

SCORPION-xisa [38]2 years ago

6 0

Answer: 15m

Step-by-step explanation:

Since 1cm is to 5m, just multiply 3 by 5 and you will get 15.

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3 years ago

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A soccer field is a rectangle 48 meters wide and 55 meters long. The coach asks players to run from one corner to the corner dia

Sindrei [870]

Answer:

Distance = 73 m

Step-by-step explanation:

Given that,

The length of a rectangle = 48 m

The width of a rectangle = 55 m

We need to find the diagonal distance. We can use the Pythagoras theorem to find it.

$d^2=48^2+55^2\\\\d=\sqrt{5329}\\\\d=73\ m$

So, the required distance is equal to 73 m.

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2 years ago

F r e e p o i n t s b o o

Katyanochek1 [597]

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3 years ago

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If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

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1 year ago

5, if r = 3?<br> what is the answer

Papessa [141]

Answer:

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3 years ago

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