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3 years ago

Find the perimeter of the isosceles triangle in simplest radical form.

Mathematics

1 answer:

neonofarm [45]3 years ago

4 0

9514 1404 393

Answer:

6 +8√5

Step-by-step explanation:

The leg length can be reduced to simpler form:

3 +√45 = 3 +√(3²·5) = 3 +3√5

The perimeter is the sum of the side lengths, so is ...

(3 +3√5) +(3 +3√5) +(2√5) = 6 +8√5 . . . perimeter

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Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

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<h3>Given information-</h3>

The given function is,

$h(x)=x^2+1$

$k(x)=x-2$

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

$(h+k)(x)=h(x)+k(x)$

$(h+k)(x)=x^2+1+x-2$

$(h+k)(2)=2^2+1+2-2$

$(h+k)(x)=5$

2)The value of the function (h-k)(3),

$(h-k)(x)=h(x)-k(x)$

$(h-k)(x)=x^2+1-x+2$

$(h-k)(3)=3^2+1-3+2$

$(h-k)(x)=9$

3) The value of the function 3h(2)+2k(3)

$3h(x)+2k(x)=3x^2+3+2x-2\times 2$

$3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2$

$3h(2)+2k(3)=17$

Hence the value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

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