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3 years ago

Help please and simplify answer in radical form if needed

Mathematics

1 answer:

deff fn [24]3 years ago

6 0

Answer:

$16\sqrt{3}\:\mathrm{units^2}$

Step-by-step explanation:

The area of an equilateral triangle with side length $s$ is equal to $\frac{\sqrt{3}}{4}s^2$ . Since the perimeter is 24 units and there are three equal sides to an equilateral triangle, each side is $24\div 3=8$ units.

Substituting in this value, we get:

$\frac{\sqrt{3}}{4}8^2=\frac{64\sqrt{3}}{4}=\boxed{16\sqrt{3}}$

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HELP I NEED HELP ASAP

lubasha [3.4K]

Answer:

Step-by-step explanation:

sum of two numbers x and y is 140

x+y=140

the difference between x(the larger number) and y( smaller number)

x-y=32

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2 years ago

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An 11-ft ladder leans against the side of a house. The bottom of the ladder is 9-ft from the side of the house. How high is the

bulgar [2K]

Answer:

6.3 ft.

Step-by-step explanation:

This requires the pythagorean theorem: a^2+b^2=c^2 where "a" and "b" are legs and c is the hypotenuse(longest side of the triangle which is opposite to the 90 degree angle.) If we fill the equation where c is the ladder length(11) and "a" is the distance from the house(), and "b" is the height, 9^2+b^2=11^2 or 81+b^2=121. This is equivalent to b^2=40. b=6.3 when rounded to the nearest tenth.

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3 years ago

A company rents water tanks shaped like cylinders. Each tank has a radius of 6 feet and a height of 2 feet. The cost is $3 per c

choli [55]

It should be 36.00 dollars

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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3 years ago

Which point keeps the set of ordered pairs a function {(1,-2).(-2,4), (5,0)}

tester [92]

B (2,1) is the answer

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2 years ago