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2 years ago

The longest side of an acute triangle measures 30 inches. The two remaining sides are congruent, but their length is

Mathematics

1 answer:

qwelly [4]2 years ago

4 0

Answer:

Step-by-step explanation:

Given an acute triangle in which the longest side measures 30 inches; and the other two sides are congruent.

Consider the attached diagram

AB=BC=x

However to be able to solve for x, we form a right triangle with endpoints A and C.

Since the hypotenuse is always the longest side in a right triangle

Hypotenuse, AC=30 Inches

Using Pythagoras Theorem

$30^2=x^2+x^2\\900=2x^2\\x^2=450\\x=\sqrt{450}\\x=21.21$ inches$

Therefore, the smallest possible perimeter of the triangle

Perimeter=2x+30

=2(21.21)+30

=42.42+30

=72.4 Inches (rounded to the nearest tenth)

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3 years ago

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

Kay [80]

Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

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solution:

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multiply both sides by (x - 3) ≠ 0

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cancel (x - 3)

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subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$

cross multiply

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subtract 4x from both sides

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combine like terms

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change the signs

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The product is 0 if one of the factors is 0. Therefore:

$x-9=0\ \vee\ x+4=0$

$x-9=0$ add 9 to both sides

$x=9\in D$

$x+4=0$ subtract 4 from both sides

$x=-4\in D$

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