What is the measure of arc bc

Mathematics

1 answer:

Hunter-Best [27]3 years ago

6 0

115 would be your answer because 180-65=115

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(-4,5) and (4,0) find the distance between each pair of points

aleksklad [387]

Answer:

9.4339 units

Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

Here we are going to use the distance formula, which is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (-4,5)$

$(x_2,y_2) = (4,0)$

Replacing these values in the distance formula

$D =\sqrt{(-4-4))^2+(5-0)^2}$

$D =\sqrt{(-8)^2+(5)^2}$

$D =\sqrt{64+25}$

$D =\sqrt{89}$

$D= 9.4339$

Hence the distance is 9.4339 units

6 0

3 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$