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Alexxandr [17]
3 years ago
12

The distance from Jacksonville to Tampa is 171 miles. The distance from Tampa to Miami is 206 miles. Use the triangle inequality

theorem to find the range for the distance from Jacksonville to Miami.
Mathematics
1 answer:
Natali [406]3 years ago
3 0
The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the two other sides. 
That means if you add up the distance from jacksonville to tampa and the distane from tampa to miami, the distance from jacksonville to miami must be less than that. 

171 miles + 206 miles = 377 miles

the distance from jacksonville to miami must be less than 377 miles 
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Katarina [22]

Answer:

y=42-x

x=42-y

Step-by-step explanation:

x=42-y

Subtract y from both sides of the equation

y=42-x

Subtract x from both sides of the equation

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A theater sells children's tickets for half the adult ticket price. if 5 adult tickets and 8 children's tickets cost a total of
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27 = 5x/2 + 8y is an equation to use to solve 
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Find the sum of the first 17 terms of the arithmetic sequence 10, 14, 18, 22, 26...
kari74 [83]
10, 14, 18 ....

notice, we get the next term by simply adding 4 to the current term, thus "4" is the "common difference, and we know that 10 is the first term.

\bf n^{th}\textit{ term of an arithmetic sequence}
\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=10\\
d=4\\
n=17
\end{cases}
\\\\\\
a_{17}=10+(17-1)(4)\implies a_{17}=10+(16)(4)
\\\\\\
a_{17}=10+64\implies a_{17}=74\\\\
-------------------------------

\bf \textit{ sum of a finite arithmetic sequence}
\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
a_1=10\\
a_{17}=74\\
n=17
\end{cases}
\\\\\\
S_{17}=\cfrac{17(10+74)}{2}\implies S_{17}=\cfrac{17(84)}{2}\implies S_{17}=714
6 0
2 years ago
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Nataly_w [17]
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8 0
3 years ago
For what values of b are the vectors [−18, b, 9] and [b, b2, b] orthogonal? (Enter your answers as a comma-separated list. If an
bearhunter [10]

Answer:

Therefore the given vectors are orthogonal for b = 0,±3.

Step-by-step explanation:

If  \vec a and  \vec b are two vectors orthogonal, then the dot product of \vec a and \vec b will be zero.

i.e \vec a. \vec b =0

If  \vec a = x_1\hat i+y_1\hat j +z_1\hat k  and \vec b = x_2\hat i+y_2\hat j +z_2\hat k

\vec a. \vec b=( x_1\hat i+y_1\hat j +z_1\hat k).( x_2\hat i+y_2\hat j +z_2\hat k)

     =x_1 x_2+y_1y_2+z_1z_2

Given two vectors are (-18,b,9) and (b,b²,b)

Let

\vec P= -18 \hat i+b\hat j +9 \hat k

and

\vec Q = b \hat i+b^2 \hat j +b\hat k

Therefore,

\vec P.\vec Q

=( -18 \hat i+b\hat j +9 \hat k).( b \hat i+b^2 \hat j +b\hat k)

=(-18).b+b.b²+9.b

= -18b+b³+9b

= b³-9b

Since \vec P and \vec Q are orthogonal. Then \vec P.\vec Q = 0.

Therefore,

b³-9b= 0

⇒b(b²-9)=0

⇒b =0 or b²=9

⇒b=0 or b =±3

Therefore the given vectors are orthogonal for b = 0,±3.

8 0
3 years ago
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