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3 years ago

The length of the leg of right isosceles triangle is a. Find the length of the hypotenuse.

1 answer:

6 0

A right isosceles triangle has two 45 degree angles. This means the side lengths are the same, so if one leg is a, the other leg is also a.

The hypotenuse of an isosceles triangle is the length os a leg multiplied by the square root of 2.

The hypotenuse would be: a√2

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Which is greater, 1/4 or 9/20

Zina [86]

9/20

You need to have the same denominator so multiply it by 5 on the bottom and top to get 5/20 then you can compare it 5/20 < 9/20

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3 years ago

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Help, i will give brainliest

andrey2020 [161]

Answer:

The answer is J

Step-by-step explanation:

If you start from Quinn's location (8,7), and move south 3 units, you get to (8,4). Then you move to the west 4 units and that gets you to (4,4).

hope this helped! :)

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3 years ago

mr yeo had 32litre of water.he used 3/8 of it to fill up a fish tank.he used 1/4 of the remaining water to water the plants. how

aleksandr82 [10.1K]

So, Mr Yeo has 32 litres of water to begin, by taking away the 3/8 (12 litres) you get 20 litres, dividing that by 4 gives you 5litres.

Your answer is then 5 Litres

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3 years ago

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Find the number that makes the ratio equivalent 1:3 9:_

dimaraw [331]

Answer:

Step-by-step explanation:

If the ratio is 1:3 multiply that by 9:9 so you effectively get 9:9x3 = 9:27. Or, you can say to yourself if B (3) is 3 times greater than A (1) then B is 3 times greater than 9. So 9x3 = 27

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3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago