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Alecsey [184]
2 years ago
15

Find the length of the missing side.

Mathematics
2 answers:
umka21 [38]2 years ago
4 0

Answer:

\displaystyle   4 \sqrt{13}in

Step-by-step explanation:

we are given a right angle triangle and we want to figure out the missing <u>hypotenuse</u><u> </u>in order to do so we can consider using<u> </u><u>Pythagoras</u><u> theorem</u> given by

\displaystyle  {a}^{2}  +  {b}^{2}  =  {c}^{2}

\displaystyle    \implies{c}^{}  =  \sqrt{ {a }^{2}  +  {b}^{2} }

let a and b be 8 and 12 respectively thus

substitute:

\displaystyle   {c}^{}  =  \sqrt{ {8 }^{2}  +  {12}^{2} }

simplify squares:

\displaystyle   {c}^{}  =  \sqrt{ 64 +  14 4}

simplify addition:

\displaystyle   {c}^{}   =  \sqrt{208}

by simplifying the redical expression we obtain:

\displaystyle   {c}^{}   = 4 \sqrt{13}

White raven [17]2 years ago
3 0

Answer:

14.4 inches

Step-by-step explanation:

Use the Pythagorean theorem: a^{2} + b^{2}

So, this will be: 8^{2} + 12^{2} = c^{2}

By simplifying we get:

c = \sqrt{208}

The missing side is: 14.4 (rounded to the tenth)

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Please help <br><br> (only #15 to do)
mote1985 [20]
Whats the question?

Sorry but can't answer if theres no question

3 0
3 years ago
Which value below is a solution to the inequality 4x - 5 &gt; 37
ZanzabumX [31]

Answer:

3

Step-by-step explanation:

The answer to the question is x > 16

This shows that x is bigger than 16

4 0
2 years ago
What does 2 (p -q) +5 (p+q) equal when p = 7 and q = -2? Answer:​
salantis [7]

Answer:

43

Step-by-step explanation:

Given

2(p - q) + 5(p + q) ← substitute p = 7 and q = - 2 into the expression

= 2(7 - (- 2)) + 5(7 + (- 2))

= 2(7 + 2) + 5(7 - 2)

= 2(9) + 5(5)

= 18 + 25

= 43

7 0
3 years ago
10. Y is the midpoint between points X and Z. If Y has a coordinate of (6,4) and Z has a coordinate of
inysia [295]

Answer 10:

<em>X </em>= (5,0)

Answer 11:

<em>AB</em> = 11.18

Answer 12:

<em>N/A</em>

Step-by-step explanation:

3 0
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}
7 0
3 years ago
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