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Zigmanuir [339]
3 years ago
7

What is the answer to this question? Write an expression in simplest form for the perimeter of a right triangle with leg lengths

of 3a to the third power and 4a to the third power.
Mathematics
1 answer:
blagie [28]3 years ago
7 0

Answer:

Step-by-step explanation:

Since the length of both legs of the right angle triangle are given, we would determine the hypotenuse, h by applying Pythagoras theorem which is expressed as

Hypotenuse² = one leg² + other leg²

Therefore,

h² = (3a)³ + (4a)³

h² = 27a³ + 64a³

h² = 91a³

Taking square root of both sides,

h = √91a³

The formula for determining the perimeter of a triangle is expressed as

Perimeter = a + b + c

a, b and c are the side length of the triangle. Therefore, the expression for the perimeter of the right angle triangle is

√91a³ + (3a)³ + (4a)³

= √91a³ + 91a³

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ALGEBRA QUESTION PLS HELPS
cluponka [151]

The value of x is –7.

Solution:

Given expression:

$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}

Let us factor x^2+4x+3.

x^2+4x+3=(x+1)(x+3)

Substitute this in the fraction.

$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

To make the denominator same, multiply and divide the first term by (x +1).

$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

Denominators are same, you can add the fractions.

$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}

Cancel the common term in the numerator and denominator.

$\frac{x+7}{x+1} \cdot \frac{1}{x+1}

Multiply the fractions.

$\frac{x+7}{(x+1)^2}

$\frac{x+7}{x^2+2x+1}

The expression is simplified to one rational expression.

Suppose the expression is equal to 0.

$\frac{x+7}{x^2+2x+1}=0

Do cross multiplication.

${x+7}=0\times (}{x^2+2x+1})

Any number or variable multiplied by 0 gives 0.

${x+7}=0

Subtract 7 from both sides of the equation.

${x+7-7}=0-7

x = –7

The value of x is –7.

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