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3 years ago

2V2 km 2 km X A) 2V5 km C) 2 km B) 2V3 km D) 4 km HELP PLEASEE

Mathematics

1 answer:

Rasek [7]3 years ago

3 0

Answer:

$x=2\sqrt3\ km$

Step-by-step explanation:

Given that,

The base of a triangle = 2 km

The perpendicular height of the triangle = $2\sqrt{2} \ km$

We need to find the value of x. It is the Hypotenuse of the triangle. It can be solved as :

$x=\sqrt{(2\sqrt{2})^2+2^2 } \\\\=2\sqrt{2+1 } \\\\=2\sqrt3\ km$

So, the value of x is equal to $2\sqrt3\ km$ .

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Ipatiy [6.2K]

Answer:

15 x 6 x 8

Step-by-step explanation:

total money is cost per lawn times number of lawns per week times number of weeks

cost per lawn is $15

number of lawns per week is 6

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so cost total money is 15 x 6 x 8 which is none of them

where did the 20 and 11 come from?

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Natasha2012 [34]

The answer is 0.50.

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Notice that in 8 the line touches the x axis so it’s corresponding probability is 0.

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2 years ago

find the equation for the line that passes through (10, 10) that has slope -1/3. give your answer in point-slope form. you do no

chubhunter [2.5K]

Y - 10 =(-1/3) ( X - 10 )

Y = ( -1/3) X + ( 40/3)

Step-by-step explanation:

Step 1:

The basic form of Slope Intercept equation is

$Y - Y_{1} = m ( X - X_{1})$

Step 2:

Points are (10,10) and m=-1/3

Y-10 = (-1/3)(X-10)

Y = ( (-X + 10 )/3) + 10

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3 years ago

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lorasvet [3.4K]

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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