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2 years ago

I really need help asap! ty!

Mathematics

1 answer:

Keith_Richards [23]2 years ago

8 0

The length of AL in the triangle is 12cm.

<h3>How to calculate the value?</h3>

From the information given, angle C = 90° and BAC is given as 2ABC.

In this case, the length of angle BC is given as 24cm. Therefore, the length of AL will be:

= 24/2

= 12..

Therefore, AL = 12cm

Learn more about triangles on:

brainly.com/question/1058720

#SPJ1

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A. The distribution of sample means of the differences will be approximately normal if there are at least 30 years of data in the sample and/or if the population of differences in winning times for all years is normal.

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3 years ago

5x^2-2x=2<br>help plsmmm

Yuki888 [10]

Answer:

Presumably you're solving for x here? Without further information we'll assume that.

With that in mind, x is approximately equal to 0.86 and -0.46

Step-by-step explanation:

Let's start by putting it in the usual ax² + bx + c format.

$5x^2- 2x = 2\\5x^2 - 2x - 2 = 0\\$

let's solve it. First we'll multiply both sides by five, making the first term a perfect square:

$25x^2 - 10x - 10 = 0\\$

Now we'll add 11 to both sides:

$25x^2 - 10x + 1 = 11\\$

Which makes the left side a perfect square:

$(5x - 1)^2 = 11$

And now we can solve for x:

$(5x - 1)^2 = 11\\\\5x - 1 = \sqrt{11} \\\\5x = 1 + \sqrt{11}\\\\x = \frac{1 + \sqrt{11}}{5}$

Note that there's no apparent way of drawing the ± symbol when editing equations, so take that + sign as actually being ±.

That gives us two answers:

$x = \frac{1 + \sqrt{11}}{5}\\\\x \approx \frac{1 + 3.32}5\\\\x \approx \frac{4.32}5\\\\x \approx 0.86$ $x = \frac{1 - \sqrt{11}}{5}\\\\x \approx \frac{1 - 3.32}5\\\\x \approx \frac{-2.32}5\\\\x \approx -0.46$

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