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3 years ago

Please help as soon as possible!!

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

The correct answer is b. The HL theorem says that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent. We know that the hypotenuse is congruent in both triangles because it is shared. We know that one leg of each triangle is congruent because they are marked with a single line.

If you need more help, comment below and I'd be happy to assist.

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What is the area of this triangle?

grin007 [14]

Answer:

plz mark brainiest✌️✌️

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3 years ago

X-1/x-2+x+3/x-4=2/(x-2).(4-x)

Varvara68 [4.7K]

The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

According to the statement

we have given that the equation and we have to prove that the given answer is a correct answer for those equivalent equation.

So, The given expression are:

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

And we have to prove the answer.

So, For this

$\frac{x-1}{x-2} +\frac{x+3}{x-4}$

$\frac{({x-1}) ({x-4}) +({x+3})({x-2})} {(x-2) (x-4)}$

Then the equation become

$\frac{x^{2} -4x -x +4 + x^{2} -2x + 3x -6 }{(x-2) (x-4)}$

Now solve it then

$2x^{2} - 4x -2 / (x-2) (x-4)$

Now take 2 common from answer then equation become

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

Hence proved.

So, The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

Learn more about equations here

brainly.com/question/2972832

#SPJ1

3 0

1 year ago

Find the minimum and maximum of f(x, y, z) = y + 4z subject to two constraints, 2x + z = 4 and x2 + y2 = 1. g

yuradex [85]

$L(x,y,z,\lambda_1,\lambda_2)=y+4z+\lambda_1(2x+z-4)+\lambda_2(x^2+y^2-1)$

$L_x=2\lambda_1+2\lambda_2 x=0\implies\lambda_1+\lambda_2x=0$
$L_y=1+2\lambda_2y=0$
$L_z=4+\lambda_1=0\implies\lambda_1=-4$
$L_{\lambda_1}=2x+z-4=0$
$L_{\lambda_2}=x^2+y^2-1=0$

$\lambda_1=-4\implies \lambda_2x=4\implies\lambda_2=\dfrac4x$
$1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12\implies8y=-x$

$x^2+y^2=1\implies (-8y)^2+y^2=65y^2=1\implies y=\pm\dfrac1{\sqrt{65}}$
$y=\pm\dfrac1{\sqrt{65}}\implies x=\mp\dfrac8{\sqrt{65}}$
$2x+z=4\implies z=4\pm\dfrac{16}{\sqrt{65}}$

We have two critical points to consider: $\left(-\dfrac8{\sqrt{65}},\dfrac1{\sqrt{65}},4+\dfrac{16}{\sqrt{65}}\right)$ and $\left(\dfrac8{\sqrt{65}},-\dfrac1{\sqrt{65}},4-\dfrac{16}{\sqrt{65}}\right)$ .

At these points, we respectively have a maximum of $16+\sqrt{65}$ and a minimum of $16-\sqrt{65}$ .

6 0

3 years ago

The numbers 1–15 are written on note cards and placed in a bag. One card will be drawn from the bag at random. 1. List the sampl

Taya2010 [7]

Yes I believe they are 1 to 15. If I'm not wrong please dont report me.

6 0

3 years ago

What is a prime number

Vitek1552 [10]

Answer:

3,5,7,11

Step-by-step explanation:

Prime numbers are special numbers that can only be divided by themselves and 1.

4 0

3 years ago