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3 years ago

The sectors of a circle graph _______ add up to 180 Degrees.Sometimes, Always, or Never?

Mathematics

1 answer:

mixas84 [53]3 years ago

3 0

The sectors of a circle graph <em><u>never</u></em> add up to 180 degrees. They always add up to 360 degrees.

Hope it helps:)

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Solve for 2x^2-4x+5=6

Softa [21]

To solve for a variable, you need to get it (x) by itself.

2x² - 4x + 5 = 6 Subtract 6 from both sides
2x² - 4x - 1 = 0 Plug this into the Quadratic Formula

a = 2 , b = -4 , c = -1
x = $\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ Plug in your values
x = $\frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-1)} }{2(2)}$ Simplify
x = $\frac{4 \pm \sqrt{16 + 8} }{4}$ Simplify
x = $\frac{4 \pm 2\sqrt{6} }{4}$ Simplify by removing a 2 from every term
x = $\frac{2 \pm \sqrt{6} }{2}$ Simplify by taking the fraction apart
x = $\frac{2}{2} \pm \frac{ \sqrt{6} }{2}$ Change $\frac{2}{2}$ to 1
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x = 1 $\pm \sqrt{ \frac{3}{2} }$

4 0

3 years ago

Is 0.5814 repeating a rational or irrational number??

goldenfox [79]

Answer:

irrational

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Chuge%7B%5Cgreen%7D%5Cfcolorbox%7Bblue%7D%7Bcyan%7D%7B%5Cbf%7B%5Cunderline%7B%5Cred%7B%5C

MariettaO [177]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's find out the gradient (Slope " m ") of line q ;

$\qquad \sf \dashrightarrow \:m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$\qquad \sf \dashrightarrow \:m = \dfrac{4 - 0}{0 - 1}$

$\qquad \sf \dashrightarrow \:m = - 4$

Now, since we already know the gradient let's find of the equation of line by using its Slope and one of the points using point slope form of line :

$\qquad \sf \dashrightarrow \:y - 4 = m(x - 0)$

$\qquad \sf \dashrightarrow \:y = mx + 4$

Now, plug in the value of gradient ~

$\qquad \sf \dashrightarrow \:y = - 4x + 4$

here we can clearly observe that, the Area under the curve can easily be represented as :

$\qquad \sf \dashrightarrow \:y < - 4x + 4$

Since, all the values of y that lies in the shaded region is smaller than the actual value of y for the corresponding values of x in the equation of line q

5 0

2 years ago

Help me with this question

Ivan

B is the answer because none of the others will be right and b is the one you should always do when to solve a rea word problem

8 0

3 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$

3 0

3 years ago