An isosceles right triangle has a hypotenuse length that can be modeled with the function h(x) = sqaure root of 2 The combined

2 answers:

8 0

The legs are defined by $L(x) = 2x$ . The Hypotenuse if defined by $h(x) = L(x)\cdot \sqrt{2}$ . The perimeter is defined by:

$p(x) = h(x)+2\cdot L(x) \\~\\p(x) = L(x)\cdot\sqrt{2} +2\cdot L(x) \\~\\p(x) = L(x)\cdot\left(\sqrt{2}+2\right) \\~\\p(x) = 2x\cdot\left(\sqrt{2}+2\right) \\~\\p(x)= 2\sqrt{2}x+4x$

D. P(x) = 2x √2 + 4x

Arturiano [62]2 years ago

7 0

B
hope this helps
have a great day

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Find the equation of the line passing through points (6,-3) and (-2,3)

Eduardwww [97]

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2 years ago

Can someone pls explain how to do this

Rama09 [41]

The polynomial A is represented as x - 2, making option A the right choice.

In the question, we are asked to find the correct representation of A, when the algebraic expression (x² + x)/(x + 3) can be rewritten as the algebraic expression A + 6/(x + 3), where A is a polynomial.

Thus, we can write an equation,

A + 6/(x + 3) = (x² + x)/(x + 3),

or, A + 6/(x + 3) - 6/(x + 3) = (x² + x)/(x + 3) - 6/(x + 3) {Subtracting 6/(x + 3) from the both sides of the equation}

or, A = (x² + x)/(x + 3) - 6/(x + 3) {Simplifying},

or, A = (x² + x - 6)/(x + 3) {Subtracting the two fractions with the same denominator (x + 3)},

or, A = (x² + 3x - 2x - 6)/(x + 3) {Mid-term factorization},

or, A = {x(x + 3) - 2(x + 3)}/(x + 3) {Taking common},

or, A = {(x - 2)(x + 3)}/(x + 3) {Taking (x + 3) common},

or, A = x - 2 {Cancelling (x + 3) from the numerator and the denominator}.

Thus, the polynomial A is represented as x - 2, making option A the right choice.

Learn more about polynomials at

brainly.com/question/11434122