Multiply and divide by the product of the denominators.
$\dfrac{-2}{x+3}+\dfrac{-4}{x-5}=\dfrac{-2(x-5)-4(x+3)}{(x+3)(x-5)}\\\\=\dfrac{-6x-2}{x^{2}-2x-15}$

6 0

3 years ago

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Wasani was playing Jeopardy. In this game of Jeopardy, players earn 50 points for a correct response and lose 50 points for an i

Nady [450]

He got two correct and two wrong.

5 0

3 years ago

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A right pyramid with a regular hexagon base has a height of 3 units If a side of the hexagon is 6 units long, then the apothem i

Setler79 [48]

Answer:

Part a) The slant height is $3\sqrt{2}\ units$

Part b) The lateral area is equal to $54\sqrt{2}\ units^{2}$

Step-by-step explanation:

we know that

The lateral area of a right pyramid with a regular hexagon base is equal to the area of its six triangular faces

$LA=6[\frac{1}{2}(b)(l)]$

where

b is the length side of the hexagon

l is the slant height of the pyramid

Part a) Find the slant height l

Applying the Pythagoras Theorem

$l^{2}=h^{2} +a^{2}$

where

h is the height of the pyramid

a is the apothem

we have

$h=3\ units$

$a=3\ units$

substitute

$l^{2}=3^{2} +3^{2}$

$l^{2}=18$

$l=3\sqrt{2}\ units$

Part b) Find the lateral area

$LA=6[\frac{1}{2}(b)(l)]$

we have

$b=6\ units$

$l=3\sqrt{2}\ units$

substitute the values

$LA=6[\frac{1}{2}(6)(3\sqrt{2})]=54\sqrt{2}\ units^{2}$

3 0

3 years ago

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