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Harlamova29_29 [7]
3 years ago
11

find the distance between the pair of points. ( square root 5, square root 2) and -3 square root 5, 5 square root 2)

Mathematics
1 answer:
Nata [24]3 years ago
7 0
Use the distance formula 

d  = sqrt ((y2-y1)^2  + (x2-x1)^2))

    =   sqrt ( (5 sqrt2 - sqrt2)^2  +  ( -3sqrt5 - sqrt5)^2)
  
   =     sqrt [( 4sqrt2)^2 + (4sqrt5)^2]

   = sqrt   (32 +  16*5)  =  sqrt 112  = sqrt16 * sqrt7  = 4 sqrt7

The answer is  4 sqrt7
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Answer:

a function of an angle, or of an abstract quantity, used in trigonometry, including the sine, cosine, tangent, cotangent, secant, and cosecant, and their hyperbolic counterparts.

Step-by-step explanation:

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2 years ago
How do you this question?
Svetlanka [38]

Answer:

i think the answer is in the question it is -4,1

Step-by-step explanation:


7 0
3 years ago
1 pt) find the average rate of change of the function over the given intervals. h(t)=\cot (t);
blondinia [14]

\text{Consider the function}\\
\\
h(t)=\cot(t) \text{ on the interval }\left [ \frac{\pi}{4}, \ \frac{3\pi}{4} \right ]\\
\\
\text{we know that the average rate of change of a funtion f(x) over the }\\
\text{interval [a, b] is given by}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this, the average rate of change of the given function is}

h_{avg}=\frac{h\left ( \frac{3\pi}{4} \right )-h\left ( \frac{\pi}{4} \right )}{\frac{3\pi}{4}-\frac{\pi}{4}}\\
\\
=\frac{\cot \left ( \frac{3\pi}{4} \right )-\cot \left ( \frac{\pi}{4} \right )}{\frac{3\pi-\pi}{4}}\\
\\
=\frac{(-1)-(1)}{\frac{2\pi}{4}}\\
\\
=\frac{-2}{\frac{\pi}{2}}\\
\\
\text{Average rate of change of function}=\frac{-4}{\pi}

6 0
3 years ago
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