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2 years ago

1. What is the measure of angle x? O 60° O 90° O 45° O 30°

Mathematics

1 answer:

True [87]2 years ago

5 0

Answer:

Angle x= 60°

Step-by-step explanation:

For the triangle HIL, you're going to add 30+90=120

Angles in a triangle add up to 180.

So, therefore,

180-120=60 so angle L must equal 60.

HJ is a straight line and angles in a straight line add up to 180.

180-90=90

Angle JIL is equal to 90.

To find y you add 90+45 which equals 135.

Again, angles in a triangle add up to 180.

180-135=45

So,

y = 45

Now we are told that IJK is a right angle and that we are given that IJL is 45. 45 is half of 90 so LJK must be 45.

To find angle JLK we must add angle L and angle y.

60+45=105

Angles in a straight line add up to 180. So,

180-105=75

75 = Angle JLK

75+45=120

Angles in a triangle add up to 180 so,

180-120= 60

Angle x= 60°

Hope this helped, if so please award me with the brainliest if possible. If you require further assistance from me comment below! :)

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Help needed plez<br><br>Ans in (ii) is 1.5

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OK first let's check the x=1.5.

$y = \frac x 6 (x^2 - 10)$

$y = 1-2x$

$1-2x = \frac x 6 (x^2 - 10)$

$6-12x = x^3 - 10 x$

$x^3 + 2x-6 = 0$

Oh my, that's called a depressed cubic, no $x^2$ term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that. Anyway, $x\approx 1.45616 \approx 1.5$ is a solution, but that's not what they're asking. They are asking us to compare

$x^3 + 2x-6 = 0$

with

$x^3 + bx^2 + cx + d = 0$

and conclude $b=0, c= 2, d=-6$

It turns out we did need all the rest of it. Save those brain cells, there's lots more math coming.

~~~~~~~~~~~~~~

I love it when the student asks for more. Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here). We use the trick of putting coefficients on the coefficients to avoid fractions.

$x^3 + 3 p x = 2 q$

has solutions

$x = \sqrt[3] { q - \sqrt{p^3 + q^2} } + \sqrt[3] {q + \sqrt{p^3 + q^2} } 

$