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3 years ago

The triangles are similar by the AA similarity postulate find the value of x ?

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

3 0

4/5 = (4 + 8)/x
4x = 5(12) = 60
x = 60/4 = 15

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Simplify 3x -5 + 25x-9

mash [69]

Answer:

28x - 14

Step-by-step explanation:

3x - 5 + 25x - 9

= 28x - 14

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How to find the area?????????????

Illusion [34]

Area of a triangle = 1/2 b•h

A = 1/2 b • h
A = 1/2 (10 cm • 9 cm)
A = 1/2 (90 cm^2)
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Neporo4naja [7]

Answer:

Radians

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Step-by-step explanation:

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3 years ago

Find all solutions to x³+3x² − 9x − 27 = 0 by factoring the equation.

SCORPION-xisa [38]

Answer:

The solutions are $x=-3,\:x=3$

Step-by-step explanation:

To factor this cubic polynomial $x^3+3x^2-9x-27$ you must:

Group the polynomial into two sections

$x^3+3x^2-9x-27=\left(x^3+3x^2\right)+\left(-9x-27\right)$

Factor out -9 from $(-9x-27)$

$(-9x-27)=-9\left(x+3\right)$

Factor out $x^2$ from $(x^3+3x^2)$

$(x^3+3x^2)=x^2\left(x+3\right)$

$x^3+3x^2-9x-27=-9\left(x+3\right)+x^2\left(x+3\right)$

Factor out common term $x+3$

$-9\left(x+3\right)+x^2\left(x+3\right)=\left(x+3\right)\left(x^2-9\right)$

$x^3+3x^2-9x-27=\left(x+3\right)\left(x^2-9\right)$

Factor $x^2-9$

$x^2-9=\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27= \left(x+3\right)\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27=\left(x+3\right)^2\left(x-3\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+3=0, \quad{x=-3}\\x-3=0, \quad{x=3}$

The solutions are

$x=-3,\:x=3$

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3 years ago

Is 12.8 greater than 12.815?

Arturiano [62]

Answer:

No,

12.8 = 12.800

Obviously, 12.800 <u>< </u>12.815.

$\infty \infty$

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