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3 years ago

What is the value of k?

Mathematics

1 answer:

seraphim [82]3 years ago

8 0

Answer:

k=2

Step-by-step explanation:

We have got a right triangle LNO with sides = m,4 and 8

By Pythagorean theorem

m^2=4^2 +8^2

Or m^2 = 80 ... i

IN triangle NOM, 4^2 +k^2 = l^2

i.e. 16+k^2 = l^2... i

In triangle LNM, m^2+l^2 = (8+k)^2

Substitute for m^2 as 80,

i.e. 80+l^2 = 64+k^2 +16k ... iii

From i, l^2 = 16+k^2

Substitute in iii

80+16+k^2 = 64+k^2 +16k

16k = 32 or k =2

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2 years ago

Solve irrational equation pls

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$\hbox{Domain:}\\
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$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
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(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
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There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
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Finally
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\boxed{\boxed{x=1}}$

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2 years ago

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kumpel [21]

ANSWER

$x=-2,y=-6$

EXPLANATION

First equation:

$y = x - 4$

Second equation:

$y = 4x + 2$

Let us equate the two equations and solve for x.

This implies that;

$4x + 2 = x - 4$

Group similar terms to obtain:

$4x - x = - 4 - 2$

Simplify:

$3x = - 6$

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Put x=-2 into any of the equations.

I will put x=-2 into the first equation:

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