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2 years ago

Determine the distance between the two points using the Pythagorean theorem.

Mathematics

1 answer:

galina1969 [7]2 years ago

6 0

Answer:

d = 10

Step-by-step explanation:

Ok, so we know that the horizontal distance between the points is 6 and the vertical distance is 8. Using the Pythagorean theorem we know that 8^2+6^2=d^2 (d is the distance between points). When solving we get

8^2+6^2=d^2

64+36=d^2

100=d^2

d=10

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3 years ago

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3 years ago

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3 years ago

Can someone please explain me this

Vlad [161]

1st way

To solve this we need to know ratio.

$- \frac{1}{5} \div \frac{1}{5} = - 1 \\$

Now we can see that odd numbers of geometric sequence are positive and even numbers are negative. 91 - odd number, so it will be positive. If ratio is equal to -1, than 91st number is

$\frac{1}{5} \\$

2nd way

Firstly, we need to find ratio.

$- \frac{1}{5} \div \frac{1}{5} = - 1 \\$

Secondly, we need to use this formula

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3 years ago

How does the graph of y = sec(x + 3) – 7 compare with the graph of y = sec(x)?

Dmitry_Shevchenko [17]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% function transformations for trigonometric functions
% templates
f(x)=Asin(Bx+C)+D
\\\\
f(x)=Acos(Bx+C)+D\\\\
f(x)=Atan(Bx+C)+D
\\\\
-------------------$

$\bf \bullet \textit{ stretches or shrinks}\\
~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\
\bullet \textit{ flips it upside-down if }A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{C}{B}\\
~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{C}{B}\textit{ is positive, to the left}\\\\
\bullet \textit{vertical shift by }D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}$

$\bf \bullet \textit{function period or frequency}\\
~~~~~~\frac{2\pi }{B}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
~~~~~~\frac{\pi }{B}\ for\ tan(\theta),\ cot(\theta)$

with that template in mind,

$\bf y=sec(\stackrel{B}{1}x+\frac{C}{3})\stackrel{D}{-7}$

the derived function has a horizontal shift of C/B or +3/1 or +3, namely 3 units to the left.

and has a vertical shift D = -7, of 7 units downwards.

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2 years ago